8.8-8.10 usaco

summary：44

没救了。。。整天刷水迟早药丸！

❤bzoj3892：

区间dp。我原来的思路是dp[i][j]表示前i个数跳过了j次，那么转移可以前k个数转移了j-1次，枚举k就好了，但是这样是错的，因为前k个数转移了j-1次，那么再k到i之间到底要在哪一步跳无法确定。于是便WA了。正确的转移应该是前i-k-1个数转移了j-k次，在i-k-1到i直接跳过去就好了。dp的转移方程要多验证是否是对的。。。！

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0,f=1;char c=getchar();
	while(!isdigit(c)){
		if(c=='-') f=-1;c=getchar();
	}
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x*f;
}
const int nmax=505;
const int inf=0x7f7f7f7f;
int dp[nmax][nmax],x[nmax],y[nmax];
int as(int x){
	return x<0?-x:x;
}
int main(){
	int N=read(),K=read();
	rep(i,1,N) x[i]=read(),y[i]=read();
	clr(dp,0x7f);dp[1][0]=0;
	rep(i,2,N) {
		for(int j=0;j<=i-2&&j<=K;j++){
			rep(k,0,j)
			  dp[i][j]=min(dp[i][j],dp[i-k-1][j-k]+as(x[i]-x[i-k-1])+as(y[i]-y[i-k-1]));
		}
	}
	printf("%d\n",dp[N][min(K,N-2)]);
}

bzoj3891：

直接三次bfs就好了。我傻叉写了三个bfs。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=40005;
const int maxn=100005;
const int inf=0x7f7f7f7f;
struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
	pt->to=u;pt->next=head[v];head[v]=pt++;
}
int dist[nmax],dis[nmax],d[nmax],B,E,P,N,M;
bool vis[nmax];
queue<int>q;
void d1(){
	clr(vis,0);vis[1]=1;
	d[1]=0;q.push(1);
	while(!q.empty()){
		int x=q.front();q.pop();
		qwq(x) if(!vis[o->to]){
			d[o->to]=d[x]+1;
			q.push(o->to);
			vis[o->to]=1;
		}
	}
}
void d2(){
	clr(vis,0);vis[2]=1;
	dis[2]=0;q.push(2);
	while(!q.empty()){
		int x=q.front();q.pop();
		qwq(x) if(!vis[o->to]){
			dis[o->to]=dis[x]+1;
			q.push(o->to);
			vis[o->to]=1;
		}
	}
}
void d3(){
	clr(vis,0);vis[N]=1;
	dist[N]=0;q.push(N);
	while(!q.empty()){
		int x=q.front();q.pop();
		qwq(x) if(!vis[o->to]){
			dist[o->to]=dist[x]+1;
			q.push(o->to);
			vis[o->to]=1;
		}
	}
}
int main(){
	B=read(),E=read(),P=read(),N=read(),M=read();
	int u,v,tmp,temp;
	rep(i,1,M) u=read(),v=read(),add(u,v);

	d1();d2();d3();
	int ans=inf;
	rep(i,1,N) ans=min(ans,B*d[i]+E*dis[i]+P*dist[i]);
	printf("%d\n",ans);
	return 0;
}

bzoj3943：

最大生成树。prim O(n2) kruskal O(eloge) 这道题边数很多用prim快了很多。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=2005;
const int maxn=4000005;
const int inf=0x7f7f7f7f;
int a[nmax],n,f[nmax];ll ans;
bool vis[nmax];
int main(){
	n=read();
	rep(i,1,n) a[i]=read();
	int tmp;
	rep(i,1,n){
		tmp=0;
		rep(j,1,n) if(!vis[j]&&f[j]>=f[tmp]) tmp=j;
		vis[tmp]=true;ans+=f[tmp];
		rep(j,1,n) f[j]=max(f[j],a[tmp]^a[j]);
	}
	printf("%lld\n",ans);
	return 0;
}
/*struct node{
	int from,to,dist;
	bool operator<(const node&rhs)const{
	  return dist>rhs.dist;}
};
node nodes[maxn];
int a[nmax],fa[nmax];
int find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
int main(){
	int n=read(),cnt=0;
	rep(i,1,n) {
		a[i]=read();
		rep(j,1,i-1){
			nodes[++cnt].from=j,nodes[cnt].to=i,nodes[cnt].dist=a[i]^a[j];
		}
	}
	sort(nodes+1,nodes+cnt+1);
	rep(i,1,n) fa[i]=i;
	int ta,tb,sum=0;ll ans=0;
	rep(i,1,cnt){
		ta=find(nodes[i].from);tb=find(nodes[i].to);
		if(ta!=tb){
			fa[ta]=tb;
			sum++;ans+=nodes[i].dist;
			if(sum==n-1) break;
		}
	}
	printf("%lld\n",ans);
	return 0;
}*/

bzoj3942：

AC自动机强上。。判断是否是子串就可以了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=1000005;
int ch[nmax][26],fail[nmax],F[nmax],T[nmax];
char s[nmax],t[nmax],S[nmax];
void insert(){
	int len=strlen(s),t=0,pt=0;
	rep(i,0,len-1){
		if(!ch[t][s[i]-'a']) ch[t][s[i]-'a']=++pt;
		t=ch[t][s[i]-'a'];
	}
	F[t]=len;
}
queue<int>q;
void getfail(){
	fail[0]=0;q.push(0);
	while(!q.empty()){
		int x=q.front();q.pop();
		rep(i,0,25) {
			if(ch[x][i]) q.push(ch[x][i]),fail[ch[x][i]]=x==0?0:ch[fail[x]][i];
			else ch[x][i]=x==0?0:ch[fail[x]][i];
		}
	}
}
int main(){
	scanf("%s",t+1);
	scanf("%s",s);insert();getfail();
	int len=strlen(t+1),top=0,x=0;
	rep(i,1,len){
		S[++top]=t[i];
		x=ch[x][t[i]-'a'];T[top]=x;
		if(F[x]) top-=F[x],x=T[top];
	}
	rep(i,1,top) printf("%c",S[i]);printf("\n");
	return 0;
}

bzoj3412：

lower_bound查找即可。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=50005;
int sum[nmax];
int main(){
	int n=read(),Q=read(),u,tmp,temp;
	sum[0]=-1;
	rep(i,1,n) u=read(),sum[i]=sum[i-1]+u;
	rep(i,1,Q){
		u=read();
		tmp=lower_bound(sum+1,sum+n+1,u)-sum;
		printf("%d\n",tmp);
	}
	return 0;
}

bzoj3431：

在枚举距离的同时枚举时间。不错的模拟！

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=10005;
double a[nmax],b[nmax];
char s[5];
int main(){
	int n,asum=0,bsum=0;scanf("%d",&n);
	double u,tmp,temp;
	rep(i,1,n){
		scanf("%s%lf",s,&u);
		if(s[0]=='T') a[++asum]=u;
		else b[++bsum]=u;
	}
	b[++bsum]=1000;
	sort(a+1,a+asum+1);
	sort(b+1,b+bsum+1);
	int v=1,now=1;
	double pre=0,dist;
	rep(i,1,bsum){
		dist=b[i]-b[i-1];
		while(pre+dist*v>a[now]&&now<=asum){
			dist-=(a[now]-pre)/v;
			pre=a[now];now++;v++;
		}
		pre+=dist*v;
		v++;
		//printf("%lf %d\n",pre,v);
	}
	int ans=pre+0.5;
	printf("%d\n",ans);
	return 0;
}

bzoj3432:

二分答案即可。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();int f=1;
	while(!isdigit(c)) {
		if(c=='-') f=-1;c=getchar();
	}
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x*f;
}
const int nmax=505;
const int inf=0x7f7f7f7f;
int map[nmax][nmax],t[nmax][nmax];
bool vis[nmax][nmax];
int N,M,cnt=0,ta,tb;

struct node{
	int x,y;
	node(int x,int y):x(x),y(y){};
};
queue<node>q;
int as(int x){
	return x<0?-x:x;
}

int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};
bool check(int mid){
	int x,y,tx,ty,res=0;
	clr(vis,0);vis[ta][tb]=1;
	q.push(node(ta,tb));
	while(!q.empty()){
		node o=q.front();q.pop();
		x=o.x,y=o.y;
		if(t[x][y]==1) res++;
		rep(i,1,4){
			tx=x+xx[i],ty=y+yy[i];
			if(tx<1||ty<1||tx>N||ty>M||as(map[tx][ty]-map[x][y])>mid||vis[tx][ty]) continue;
			q.push(node(tx,ty));vis[tx][ty]=1;
		}
	}
	return res==cnt;
}
int main(){
	N=read(),M=read();
	rep(i,1,N) rep(j,1,M) map[i][j]=read();
	rep(i,1,N) rep(j,1,M) {
		t[i][j]=read();
		if(t[i][j]==1) cnt++,ta=i,tb=j;
	}
	int l=0,r=inf,mid,ans=0;
	while(l<=r){
		mid=(l+r)>>1;
		if(check(mid)) ans=mid,r=mid-1;
		else l=mid+1;
	}
	printf("%d\n",ans);
	return 0;
}

bzoj3410：

求最多能用多少个不相交的区间覆盖线段。贪心：按结束点排序，放不了就不要。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=50005;
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
struct node{
	int l,r;
	bool operator<(const node&rhs)const{
	  return r<rhs.r;}
};
node nodes[nmax];
int main(){
	int N=read(),u,v;
	rep(i,1,N) nodes[i].l=read(),nodes[i].r=read();
	sort(nodes+1,nodes+N+1);
	int ans=0,now=-1;
	rep(i,1,N) {
		if(nodes[i].l>=now) ans++,now=nodes[i].r;
	}
	printf("%d\n",ans);
	return 0;
}

❤bzoj3404：

对于游戏问题不怎么懂。。。可以先预处理出每个数是必胜/必败，该状态是必胜的当后续状态有状态是必败的，否则就是必败的。因为在预处理的时候第一次出现必败的情况他的后续状态都是必胜的，所以每个数都只有必胜或必败两种情况。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=1000005;
int ans[nmax];
int main(){
	clr(ans,0);
	rep(i,1,9) ans[i]=1;
	int maxn,minn,tmp,temp;
	rep(i,10,nmax){
		maxn=-1,minn=10;
		tmp=i;
		while(tmp){
			temp=tmp%10;
			if(temp){
				maxn=max(maxn,temp);
				minn=min(minn,temp);
			}
			tmp/=10;
		}
		if(maxn!=-1) ans[i]|=(!ans[i-maxn]);
		if(minn!=10) ans[i]|=(!ans[i-minn]);
	}
	int G;scanf("%d",&G);
	rep(i,1,G){
		scanf("%d",&tmp);
		if(ans[tmp]) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}

bzoj3433：

贪心。按结束点排序，优先进入较长的那个队。好神的贪心！

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=155;
struct node{
	int l,r;
	bool operator<(const node&rhs)const{
	  return r<rhs.r;}
};
node e[nmax];
int main(){
	int n;scanf("%d",&n);
	rep(i,1,n){
		scanf("%d%d",&e[i].l,&e[i].r);
		e[i].r--;
	}
	sort(e+1,e+n+1);
	int now=-1,pre=-1,ans=0;
	rep(i,1,n){
		if(e[i].l>now) ans++,now=e[i].r;
		else if(e[i].l>pre) ans++,pre=e[i].r;
		if(pre>now) swap(now,pre);
	}
	printf("%d\n",ans);
	return 0;
}

bzoj3403：

模拟。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
const int nmax=200005;
int a[nmax];
char s[5],ch[5];
int main(){
	int N,l=100001,r=100000,cnt=0,tmp;
	scanf("%d",&N);
	rep(i,1,N){
		scanf("%s%s",s,ch);
		if(s[0]=='A'){
			cnt++;
			if(ch[0]=='L') a[--l]=cnt;
			else a[++r]=cnt;
		}else{
			scanf("%d",&tmp);
			if(ch[0]=='L') l+=tmp;
			else r-=tmp;
		}
	}
	rep(i,l,r) printf("%d\n",a[i]);
	return 0;
}

bzoj3402：

边权为1，我傻叉还打了个dijkstra。。应该bfs一遍就好了吧。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=20005;
const int maxn=100005;
const int inf=0x7f7f7f7f;

struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
	pt->to=u;pt->next=head[v];head[v]=pt++;
}

struct node{
	int x,dist;
	node(int x,int dist):x(x),dist(dist){};
	bool operator<(const node&rhs)const{
		  return dist>rhs.dist;
	}
};
priority_queue<node>q;
int dist[nmax];
void dijkstra(int s){
	q.push(node(s,0));
	clr(dist,0x7f);dist[s]=0;
	int x,d;
	while(!q.empty()){
		node o=q.top();q.pop();
		x=o.x;d=o.dist;
		if(d!=dist[x]) continue;
		qwq(x) if(dist[o->to]>d+1){
			dist[o->to]=d+1;
			q.push(node(o->to,dist[o->to]));
		}
	}
}
int main(){
	int N=read(),M=read(),u,v;
	rep(i,1,M){
		u=read(),v=read();
		add(u,v);
	}
	dijkstra(1);
	int ans=0;
	rep(i,1,N) ans=max(ans,dist[i]);
	int res,cnt=0;
	dwn(i,N,1) if(dist[i]==ans) res=i,cnt++;
	printf("%d %d %d\n",res,ans,cnt);
	return 0;
}

bzoj3401：

又是单调栈。。。usaco相似题怎么那么多

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=1e5+5;
int a[nmax],s[nmax],ans[nmax];
int main(){
	int N=read(),r=0;
	rep(i,1,N){
		a[i]=read();
		while(r&&a[s[r]]<a[i]) ans[s[r]]=i,r--;
		s[++r]=i;
	}
	rep(i,1,N) printf("%d\n",ans[i]);
	return 0;
}

bzoj3400：

dp。可以压掉一维。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=1005;
const int maxn=2005;
const int mod=100000000;
int dp[maxn][nmax];
int main(){
	int N=read(),F=read(),u;
	u=read();dp[1][u%F]=1;dp[1][0]+=1;
	rep(i,2,N){
		u=read();
		rep(j,0,F-1) {
			dp[i][j]=(dp[i-1][j]+dp[i][j])%mod;
			dp[i][(j+u)%F]=(dp[i-1][j]+dp[i][(j+u)%F])%mod;
		}
	}
	printf("%d\n",dp[N][0]-1);
	return 0;
}//其实可以压掉一维。。如下。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=1005;
const int mod=100000000;
int dp[nmax];
int pre[nmax];
int main(){
	int N=read(),F=read(),u,v;
	u=read();pre[u%F]=1;pre[0]++;
	rep(i,2,N){
		u=read();
		rep(j,0,F-1) {
			dp[j]=(dp[j]+pre[j])%mod;
			v=(j+u)%F;
			dp[v]=(pre[j]+dp[v])%mod;
		}
		rep(j,0,F-1) pre[j]=dp[j],dp[j]=0;
	}
	printf("%d\n",pre[0]-1);
	return 0;
}

bzoj3392：

bfs。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=1005;
const int maxn=50005;
const int inf=0x7f7f7f7f;

struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
}

int q[nmax],dist[nmax];
int main(){
	int N=read(),T=read(),u,v;
	rep(i,1,N) {
		u=read(),v=read();
		add(u,v);
	}
	int l=1,r=1;q[1]=1;
	while(l<=r){
		u=q[l];l++;
		qwq(u) if(!dist[o->to]){
			dist[o->to]=dist[u]+1;
			q[++r]=o->to;
			if(o->to==T) {
				printf("%d\n",dist[o->to]+1);
				return 0;
			}
		}
	}
	printf("-1\n");
	return 0;
}

bzoj3016：

括号序列！以前看过然后不会。。。贪心！！！

#include<cstdio>
int main(){
	char c;
	int ans=0,cnt=0;
	while(1){
		c=getchar();
		if(c=='\n') break;
		if(c=='(') cnt++;
		else if(c==')'){
			cnt--;
			if(cnt<0) cnt+=2,ans++;
		}
	}
	printf("%d\n",ans+(cnt>>1));
	return 0;
}

bzoj2019：

直接将点的权值装换到边上，然后判断是否有正环（-），spfa就好了。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<deque>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=225;
const int maxn=1005;
const int inf=0x7f7f7f7f;

struct edge{
	int to,dist;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
}

queue<int>q;
int dist[nmax],in[nmax];
bool inq[nmax];
bool spfa(int s,int D,int N){
	clr(dist,0);dist[s]=-D;
	clr(inq,0);in[s]=1;
	q.push(s);
	while(!q.empty()){
		int x=q.front();q.pop();inq[x]=0;
		qwq(x) if(dist[o->to]>dist[x]+o->dist){
			dist[o->to]=dist[x]+o->dist;
			if(!inq[o->to]){
				in[o->to]++;
				if(in[o->to]>N) return 0;
				q.push(o->to);inq[o->to]=1;
			}
		}
	}
	return 1;
}

int main(){
	int D=read(),M=read(),N=read(),P=read(),S=read(),u,v,d;
	rep(i,1,M) {
		u=read();v=read(),add(u,v,-D);
	}
	rep(i,1,P) {
		u=read();v=read();d=read();
		add(u,v,d-D);
	}
	if(spfa(S,D,N)) {
		int ans=0;
		rep(i,1,N) ans=max(ans,-dist[i]);
		printf("%d\n",ans);
	}else printf("-1\n");
	return 0;
}

❤bzoj2101：

神dp。。。然后可以枚举间距巧妙压掉一维。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=5005;
int f[nmax],a[nmax],sum[nmax];
int main(){
	int N=read();
	rep(i,1,N) a[i]=read(),sum[i]=sum[i-1]+a[i];
	rep(i,0,N-1) rep(j,1,N-i) {
		f[j]=max(sum[j+i]-sum[j-1]-f[j],sum[j+i]-sum[j-1]-f[j+1]);
	}
	printf("%d\n",f[1]);
	return 0;
}

bzoj2100：

两遍dijkstra就好了。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
const int nmax=100005;
const int maxn=400005;
const int inf=0x7f7f7f7f;
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
struct edge{
	int to,dist;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
	pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}
struct node{
	int x,dist;
	node(int x,int dist):x(x),dist(dist){};
	bool operator<(const node&rhs)const{
	  return dist>rhs.dist;}
};
priority_queue<node>q;
int dist[nmax];
void dijkstra(int s){
	clr(dist,0x7f);dist[s]=0;
	q.push(node(s,0));
	while(!q.empty()){
		node o=q.top();q.pop();
		int x=o.x,d=o.dist;
		if(dist[x]!=d) continue;
		qwq(x) if(dist[o->to]>d+o->dist)
		  dist[o->to]=d+o->dist,q.push(node(o->to,dist[o->to]));
	}
}
int main(){
	int N=read(),M=read(),s=read(),a=read(),b=read();
	rep(i,1,N){
		int u=read(),v=read(),d=read();
		add(u,v,d);
	}
	dijkstra(a);
	int ans=dist[s]+dist[b];
	dijkstra(b);
	ans=min(ans,dist[s]+dist[a]);
	printf("%d\n",ans);
	return 0;
}

❤bzoj2017：

神dp！！！用了和2101差不多的方法，还更神的压掉一维。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=2005;
int a[nmax],sum[nmax],f[nmax][nmax];
int main(){
	int n=read();
	dwn(i,n,1) a[i]=read();
	rep(i,1,n) sum[i]=sum[i-1]+a[i];
	rep(i,1,n) rep(j,1,n){
		f[i][j]=f[i][j-1];
		if(i>=j+j)
		  f[i][j]=max(f[i][j],sum[i]-f[i-j-j][j+j]);
		if(i>=j+j-1)
		  f[i][j]=max(f[i][j],sum[i]-f[i-j-j+1][j+j-1]);
	}
	printf("%d\n",f[n][1]);
	return 0;
}

bzoj1673：

利用性质用dfs写背包dp（加了一个剪枝；

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define ll long long
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=1005;
ll a[nmax],sum[nmax],ans=0,N,C;
void dfs(int cur,ll x){
	if(sum[cur-1]+x<=C) {
		ans=max(ans,sum[cur-1]+x);
		return ;
	}
	ans=max(ans,x);
	dwn(i,cur-1,1){
		x+=a[i];
		if(x<=C) dfs(i,x);
		x-=a[i];
	}
}
int main(){
	scanf("%lld%lld",&N,&C);
	rep(i,1,N) scanf("%lld",&a[i]),sum[i]=sum[i-1]+a[i];
	dfs(N+1,0);
	printf("%d\n",ans);
	return 0;
}

bzoj2016：

裸的二分答案，输出方案我原来那样子写在最后还更新了一下答案！最后还更新了答案！！！没救了。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
ll read(){
	ll x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=50005;
ll a[nmax],res[nmax],N,D;
/*bool check(ll x){
	ll cur=0,tmp=0;
	rep(i,1,D){
		while(tmp<x&&cur<=N) tmp+=a[++cur],res[cur]=i;
		if(cur>N) return false;
		tmp/=2;
	}
	rep(i,cur+1,N) res[i]=D;
	return true;
}*/
bool check(ll x){
	ll cur=0,tmp=0;
	rep(i,1,D){
		while(tmp<x&&cur<=N) tmp+=a[++cur];
		if(cur>N) return false;
		tmp/=2;
	}
	return true;
}
void get(ll x){
	ll cur=0,tmp=0;
	rep(i,1,D){
		while(tmp<x) tmp+=a[++cur],res[cur]=i;
		tmp/=2;
	}
	rep(i,cur+1,N) res[i]=D;
}
int main(){
	N=read(),D=read();
	ll l=0,r=0;
	rep(i,1,N) a[i]=read(),r+=a[i];
	ll mid,ans;
	while(l<=r){
		mid=(l+r)>>1;
		if(check(mid)) ans=mid,l=mid+1;
		else r=mid-1;
	}
	get(ans);
	printf("%lld\n",ans);
	rep(i,1,N) printf("%lld\n",res[i]);
	return 0;
}
/*
5 5
10
40
13
22
7
*/

bzoj2015：

dijkstra。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=50005;
const int maxn=200005;
const int inf=0x7f7f7f7f;
struct edge{
	int to,dist;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
	pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}

struct node{
	int x,dist;
	node(int x,int dist):x(x),dist(dist){};
	node(){};
	bool operator<(const node&rhs)const{
	  return dist>rhs.dist;}
};
priority_queue<node>q;
int dist[nmax];
void dijkstra(){
	clr(dist,0x7f);dist[1]=0;
	q.push(node(1,0));
	int x,d;
	while(!q.empty()){
		node o=q.top();q.pop();
		x=o.x,d=o.dist;
		if(dist[x]!=d) continue;
		qwq(x) if(dist[o->to]>dist[x]+o->dist){
			dist[o->to]=dist[x]+o->dist;
			q.push(node(o->to,dist[o->to]));
		}
	}
}
int main(){
	int N=read(),M=read(),B=read(),u,v,d;
	rep(i,1,M){
		u=read(),v=read(),d=read();
		add(u,v,d);
	}
	dijkstra();
	rep(i,1,B){
		u=read(),v=read();
		printf("%d\n",dist[u]+dist[v]);
	}
	return 0;
}

bzoj2014：

数据大唬人啊。。。然并卵。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
ll read(){
	ll _x=0;char _c=getchar();
	while(!isdigit(_c)) _c=getchar();
	while(isdigit(_c)) _x=_x*10+_c-'0',_c=getchar();
	return _x;
}
const int nmax=100005;
struct node{
	ll w,c;
	bool operator<(const node&rhs)const{
	  return c<rhs.c;}
};
node nodes[nmax];
int main(){
	ll N=read(),m=read();
	rep(i,1,N) nodes[i].c=read(),nodes[i].w=read();
	sort(nodes+1,nodes+N+1);
	ll ans=0,tmp;
	rep(i,1,N){
		tmp=min(m/nodes[i].c,nodes[i].w);
		ans+=tmp;m-=nodes[i].c*tmp;
	}
	printf("%lld\n",ans);
	return 0;
}

bzoj1689：

在枚举水坑的时候枚举木板就可以了

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int _x;char _c;
int read(){
	_x=0;_c=getchar();
	while(!isdigit(_c)) _c=getchar();
	while(isdigit(_c)) _x=_x*10+_c-'0',_c=getchar();
	return _x;
}
const int nmax=10005;
struct node{
	int l,r;
	bool operator<(const node&rhs)const{
	  return l<rhs.l||l==rhs.l&&r<rhs.r;}
};
node nodes[nmax];
int main(){
	int N=read(),L=read();
	rep(i,1,N) nodes[i].l=read()+1,nodes[i].r=read();
	sort(nodes+1,nodes+N+1);

	int cnt=(nodes[1].r-nodes[1].l)/L+1,cur=nodes[1].l+cnt*L-1,tmp;
	rep(i,2,N){
		if(cur>=nodes[i].r) continue;
		if(nodes[i].l>cur){
			tmp=(nodes[i].r-nodes[i].l)/L+1;
			cur=nodes[i].l+tmp*L-1;cnt+=tmp;
		}else{
			tmp=(nodes[i].r-cur-1)/L+1;
			cur+=tmp*L;cnt+=tmp;
		}
	}
	printf("%d\n",cnt);
	return 0;
}

bzoj1688：

N很小很明显可以状压。然后位运算加括号加括号不然总是神奇的优先级啊。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int _x;char _c;
int read(){
	_x=0;_c=getchar();
	while(!isdigit(_c)) _c=getchar();
	while(isdigit(_c)) _x=_x*10+_c-'0',_c=getchar();
	return _x;
}
const int nmax=1005;
int N,D,K,a[nmax],t[16];
bool pd(int x){
	int cnt=0;
	while(x) cnt+=x&1,x>>=1;
	return cnt<=K;
}
int main(){
	N=read(),D=read(),K=read();
	int u,v;
	rep(i,1,16) t[i]=1<<(i-1);
	rep(i,1,N){
		u=read();
		rep(j,1,u) v=read(),a[i]|=t[v];
	}
	int ans=0,sum,tmp=(1<<D)-1;
	rep(i,0,tmp) if(pd(i)){
		sum=0;
		rep(j,1,N) if((i|a[j])==i) sum++;
		ans=max(ans,sum);
	}
	printf("%d\n",ans);
	return 0;
}
/*
6 3 2
0
1 1
1 2
1 3
2 2 1
2 2 1
*/

bzoj1684：

暴力！模拟！

#include<cstdio>
double as(double x){
	return x<0?-x:x;
}
int main(){
	int N,D;
	scanf("%d%d",&N,&D);
	double ans=2.0,tmp=N*1.0/D;
	int ta,tb,temp;
	for(int i=1;i<=32767;i++){
		temp=tmp*i;
		if(as(tmp-temp*1.0/i)<ans&&i*N!=D*temp) ans=as(tmp-temp*1.0/i),ta=temp,tb=i;
		if(as(tmp-(temp+1)*1.0/i)<ans&&i*N!=D*(temp+1)) ans=as(tmp-(temp+1)*1.0/i),ta=temp+1,tb=i;
	}
	printf("%d %d\n",ta,tb);
	return 0;
}

bzoj1682：

二分答案+spfa。。。。我。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
const int nmax=2005;
const int maxn=20005;
const int inf=0x7f7f7f7f;
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}

struct edge{
	int to,dist;
	bool flag;
	edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
	pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}

int N,M;
int q[nmax];
bool inq[nmax];
int dist[nmax];
bool spfa(){
	clr(inq,0);inq[1]=1;
	int x,l=1,r=1;q[1]=1;
	clr(dist,0x7f);dist[1]=0;
	while(l<=r){
		x=q[l];l++;
		qwq(x) if(o->flag&&dist[o->to]>dist[x]+o->dist){
			dist[o->to]=dist[x]+o->dist;
			if(!inq[o->to]) q[++r]=o->to,inq[o->to]=1;
		}
	}
	rep(i,1,N) if(dist[i]==inf) return false;
	return true;
}
bool check(int x){
	rep(i,1,N) qwq(i)
	  if(o->dist<=x) o->flag=true;
	  else o->flag=false;
	if(spfa()) return true;
	return false;
}

int main(){
	N=read(),M=read();
	int u,v,d,omax=0;
	rep(i,1,M){
		u=read(),v=read(),d=read();add(u,v,d);
		omax=max(omax,d);
	}
	int l=0,r=omax,mid,ans=0;
	while(l<=r){
		mid=(l+r)>>1;
		if(check(mid)) ans=mid,r=mid-1;
		else l=mid+1;
	}
	printf("%d\n",ans);
	return 0;
}

bzoj1649：

找出每一段使用的次数就可以了。（分成两半想。。不然会乱死。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define ll long long
const int nmax=10005;
int a[nmax];
int main(){
	int n;scanf("%d",&n);
	rep(i,1,n) scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	ll ans=0;
	rep(i,1,n-1) ans+=(ll)i*(n-i)*(a[i+1]-a[i]);
	printf("%lld\n",ans<<1);
	return 0;
}

bzoj1677：

背包dp。。。我。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int a[25],dp[1000005];
int main(){
	int n;scanf("%d",&n);
	rep(i,0,25){
		if((1<<i)>n) break;
		a[++a[0]]=(1<<i);
	}
	dp[0]=1;
	rep(i,1,a[0]) rep(j,a[i],n)
	  if((dp[j]+=dp[j-a[i]])>=1000000000)
	    dp[j]-=1000000000;
	printf("%d\n",dp[n]);
	return 0;
}

bzoj1671：

两遍bfs就可以了（看清题意看清题意。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=1005;
const int maxn=1000005;
int map[nmax][nmax],oa[maxn],ob[maxn];
int dist[nmax][nmax],dis[nmax][nmax];
bool vis[nmax][nmax];
int q[maxn],Q[maxn];
int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};
int main(){
	int n,m,i,j,sa,sb,ta,tb,cnt=0;
	m=read(),n=read();
	rep(i,1,n) rep(j,1,m){
		map[i][j]=read();
		if(map[i][j]==2) sa=i,sb=j;
		else if(map[i][j]==3) ta=i,tb=j;
		else if(map[i][j]==4) oa[++cnt]=i,ob[cnt]=j;
	}

	int x,y,tx,ty,l,r;
	q[l=r=1]=sa;Q[l]=sb;
	vis[sa][sb]=1;map[ta][tb]=1;
	clr(dist,0x3f);dist[sa][sb]=0;
	while(l<=r){
		x=q[l];y=Q[l];l++;
		rep(i,1,4){
			tx=x+xx[i],ty=y+yy[i];
			if(tx<1||ty<1||tx>n||ty>m||map[tx][ty]==1||vis[tx][ty]) continue;
			vis[tx][ty]=1;dist[tx][ty]=dist[x][y]+1;
			q[++r]=tx;Q[r]=ty;
		}
	}

	q[l=r=1]=ta;Q[l]=tb;
	clr(vis,0);vis[ta][tb]=1;
	clr(dis,0x3f);dis[ta][tb]=0;
	while(l<=r){
		x=q[l];y=Q[l];l++;
		rep(i,1,4){
			tx=x+xx[i],ty=y+yy[i];
			if(tx<1||ty<1||tx>n||ty>m||map[tx][ty]==1||vis[tx][ty]) continue;
			vis[tx][ty]=1;dis[tx][ty]=dis[x][y]+1;
			q[++r]=tx;Q[r]=ty;
		}
	}

	int ans=0x7fffffff;
	rep(i,1,cnt) ans=min(ans,dist[oa[i]][ob[i]]+dis[oa[i]][ob[i]]);
	printf("%d\n",ans);
	return 0;
}

bzoj1665：

转换成最短路就可以了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=10005;
const int maxn=2000005;
const int inf=0x7f7f7f7f;
struct edge{
	int to,dist;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
	pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}
int H,N,s,t;
queue<int>q;
int dist[nmax],x[nmax],y[nmax];
bool inq[nmax];
void spfa(){
	clr(dist,0x7f);dist[0]=0;
	clr(inq,false);q.push(0);
	while(!q.empty()){
		int x=q.front();q.pop();inq[x]=false;
		qwq(x) if(dist[o->to]>dist[x]+o->dist){
			dist[o->to]=dist[x]+o->dist;
			if(!inq[o->to]) q.push(o->to),inq[o->to]=true;
		}
	}
	printf("%d\n",dist[t]-1);
}
int main(){
	H=read(),N=read(),s=0,t=N+1;
	rep(i,1,N) {
		x[i]=read(),y[i]=read();
		if(y[i]<=1000) add(s,i,1);
		if(H-y[i]<=1000) add(i,t,1);
		rep(j,1,i-1) if((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])<=1000000) add(i,j,1);
	}
	spfa();
	return 0;
}

bzoj1672：

更区间覆盖差不多。。改一下就可以了。挖坑！有比O(n2)的方法！

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=10005;
const int inf=0x7f7f7f7f;
struct node{
	int l,r,c;
	bool operator<(const node&rhs)const{
	  return l<rhs.l||l==rhs.l&&r<rhs.r;}
};
node nodes[nmax];
int dp[nmax];
int main(){
	int n=read(),s=read(),t=read();
	rep(i,1,n) {
		node &o=nodes[i];
		o.l=read(),o.r=read(),o.c=read();
	}
	sort(nodes+1,nodes+n+1);
	if(nodes[1].l>s) {
		printf("-1\n");return 0;
	}
	int ans=inf;
	clr(dp,0x7f);
	rep(i,1,n) {
		if(nodes[i].l<=s) dp[i]=nodes[i].c;
		rep(j,1,i-1)
		  if(dp[j]!=inf&&nodes[i].l<=nodes[j].r+1)
		    dp[i]=min(dp[i],dp[j]+nodes[i].c);
		if(nodes[i].r>=t) ans=min(ans,dp[i]);
	}
	if(ans==inf) ans=-1;
	printf("%d\n",ans);
	return 0;
}
/*
3 0 4
0 2 3
3 4 2
0 0 1
*/

bzoj1664：

同前一题

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=10005;
struct node{
	int l,r;
	bool operator<(const node&rhs)const{
	  return l<rhs.l||l==rhs.l&&r<rhs.r;}
};
node nodes[nmax];
int dp[nmax];
int main(){
	int n=read();
	rep(i,1,n) {
		node &o=nodes[i];
		o.l=o.r=read(),o.r+=read()-1;
	}
	sort(nodes+1,nodes+n+1);
	int ans=1;
	rep(i,1,n) {
		dp[i]=1;
		rep(j,1,i-1)
		  if(nodes[i].l>nodes[j].r)
		    dp[i]=max(dp[i],dp[j]+1);
		ans=max(ans,dp[i]);
	}
	printf("%d\n",ans);
	return 0;
}

bzoj1662：

！！！打表第一个元素为0，打表第一个元素为0，打表第一个元素为0！！！哦一开始的表还打不完整，可以同时打两个表然后一个输出编号。那么就可以判断打的完不完整啦。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;i++)
const int a[]={0,98891,94198,95493,107314,87316,114008,102469,74236,81159,56924,101809,143253,116151,126639,96187,88604,109609,92456,80718,69276,40647,162224,131672,118369,107427,85088,115982,100747,74503,80467,56901,90275,106606,73627,91132,57627,60303,68604,56911,50224,36158,25284,182796,162761,151564,143141,122235,147485,137996,112969,116985,92987,118295,143177,110808,128802,93230,92122,107984,92379,83907,66388,47099,159532,131378,118327,106305,80363,120639,98568,77976,74374,56700,94413,105948,69499,91945,57927,63850,69546,52748,51296,34407,34730,181177,162526,150415,143712,118099,152204,135651,114195,112068,92595,123204,142014,106644,131594,93792,96040,107083,87035,86134,63474,54978,155225,131401,118634,106767,75572,124211,96292,79755,70148,56243,97011,107586,70853,90733,57056,65433,69447,48145,53132,32025,43582,154467,130916,118508,106264,73367,126794,94239,80429,68353,57400,100958,106012,71604,87519,57572,66722,68874,46056,53415,32182,44814,116453,92806,81294,68677,38324,91605,56472,49323,38518,26743,75081,65806,41655,49212,28901,40659,36969,20942,28038,12701,48768,191146,183448,175274,170660,144282,181071,163294,151722,143637,128215,155597,170036,144933,157346,130629,131128,143583,117216,126583,96530,94970,175318,161924,150364,142014,107487,162841,131772,118795,106712,86938,137207,139655,110863,119575,93918,101314,106872,74787,90574,58392,77998,173327,161517,150494,143557,108365,161790,131378,118559,107306,83653,140360,137876,112471,118470,93274,102168,105843,73340,91392,57145,79091,142486,131788,119509,105816,73165,126604,93786,81687,68979,45646,116163,99891,76912,76111,56409,71596,68496,38005,56106,28712,75237,171313,162188,152366,141097,111110,158835,131071,118011,106090,80018,147115,136526,113900,112986,92967,103988,106200,68484,93243,57896,85415,143704,127431,121422,103119,77086,121899,92512,82512,69119,42360,122159,95976,79714,71293,56364,72488,68224,39380,55185,29396,78322,143244,125731,122087,101509,79241,119666,93681,81638,68817,40477,124806,95241,80022,69266,56160,75022,68634,40038,52425,28887,72408,107048,83085,87067,61933,52320,78065,56852,48415,37218,17262,90792,57131,48172,38127,28227,47200,35388,19556,23969,12007,91618,188003,179364,178374,167810,148695,177440,161855,151351,142818,108805,179486,162007,151810,143741,129537,141233,141251,109889,123713,93373,118608,170818,154340,156776,136190,118835,149288,130700,117473,105857,68125,162887,132023,118909,107313,87337,114058,102545,74125,81355,56783,109233,171190,152382,157609,134608,121075,146531,130574,116961,106606,71127,162240,131756,118473,107266,85298,115808,100887,74530,80460,56937,104744,143028,113112,128214,94751,90628,107060,92723,83137,68117,43228,127887,93540,82276,68663,46915,83876,62214,45435,42578,28802,111407,170911,147903,159816,131582,124710,143681,131020,119783,105004,74840,159737,131288,118327,106445,80258,120527,98827,77829,74534,56666,111270,142384,108398,129294,93814,95121,107398,89760,84594,65532,49355,123350,92952,80924,69825,43486,87894,59398,46563,39612,28742,109612,142175,106368,131638,94003,95864,107104,87175,86044,63529,52732,120317,93533,82220,68803,41452,88912,58860,47529,37969,28687,92365,107623,70818,90839,57161,65201,69588,48279,53095,32055,35207,79780,56684,48851,37003,17741,55657,29110,24270,16366,12188,127716,170972,144090,160466,131687,129365,143246,121634,123066,100633,86300,152593,130881,117493,106712,71152,128330,92901,81592,70190,54772,124627,141867,109152,125487,93093,98756,106214,80363,89123,60436,62379,113484,93274,81246,68242,37246,91936,58406,49636,37915,25000,125906,141143,110126,122782,92694,99695,107349,77610,90055,59005,65659,111535,92462,80994,67849,38982,93037,57378,49229,37299,23948,108481,101761,74481,80118,57538,69760,68853,40959,55474,29621,46335,69908,56198,49191,37488,20207,53469,28961,24876,16726,8018,134641,137953,112241,118736,93113,102168,106032,73318,91371,57202,71593,105908,94057,83424,67403,43061,87281,57217,48837,37410,21019,114520,100017,76814,76343,56423,71449,68601,38040,56085,28742,50846,70014,54232,51212,34234,24847,48630,28916,24419,16474,7466,118105,98168,77949,73627,56606,71930,68299,37034,56985,29726,52489,69888,51438,52150,33652,26250,46540,28742,24988,16620,6629,87869,58665,47562,38597,28732,43746,36970,18455,27300,12200,34405,37536,23384,27054,13648,13716,20914,12052,10145,5787,1649,181212,195180,188899,189346,182615,181348,188064,171340,183040,162686,159299,188058,181141,177205,167596,148246,178641,162701,151002,142596,111997,185183,184527,174229,170777,163010,164492,170904,144149,159881,131637,135607,171039,155170,156173,137655,117048,151608,130195,118445,106264,70098,188817,182976,175701,171305,162023,165413,170049,144814,158343,131232,137303,170782,154417,156707,136385,118724,149274,130840,117423,105837,68295,183061,163196,151645,143583,126716,143829,140518,110365,121347,92657,114542,143379,116397,126554,96166,88408,109883,92609,80761,68911,40398,194177,183680,175565,171120,159238,167988,168828,145597,154708,131399,140921,171039,149534,159445,133395,122791,144192,1299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int t[2];
bool pd(int x){
	t[0]=t[1]=0;
	while(x) ++t[x&1],x>>=1;
	return t[0]>=t[1];
}
int cal(int x){
	int tmp=x/200000,ans=0;
	rep(i,1,tmp) ans+=a[i];
	rep(i,x/200000*200000+1,x) if(pd(i)) ans++;
	return ans;
}
int main(){
	int n,m;scanf("%d%d",&n,&m);
	printf("%d\n",cal(m)-cal(n-1));
	return 0;
}

bzoj1661：

O（n4）暴力枚举

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=105;
int map[nmax][nmax];
char s[nmax];
int main(){
	int n;scanf("%d",&n);
	rep(i,1,n){
		scanf("%s",s+1);
		rep(j,1,n)
		  if(s[j]=='J') map[i][j]=2;
		  else if(s[j]=='*') map[i][j]=1;
		  else map[i][j]=0;
	}
	int x,y,ans=0;
	rep(i,1,n) rep(j,1,n) if(map[i][j]==2){
		rep(k,1,n) rep(t,1,n) if(map[k][t]){
			if(k==i&&t==j) continue;
			x=i-k,y=j-t;
			if(x*x+y*y<=ans) continue;
			if(i+y<=n&&k+y<=n&&j-x<=n&&t-x<=n&&i+y>0&&k+y>0&&j-x>0&&t-x>0&&map[i+y][j-x]==2&&map[k+y][t-x]==2)
			  ans=x*x+y*y;
			x=k-i,y=t-j;
			if(i+y<=n&&k+y<=n&&j-x<=n&&t-x<=n&&i+y>0&&k+y>0&&j-x>0&&t-x>0&&map[i+y][j-x]==2&&map[k+y][t-x]==2)
			  ans=x*x+y*y;
		}
	}
	printf("%d\n",ans);
	return 0;
}
/*
6

J*J***

******

J***J*

******

******

******
*/

bzoj1660：

又是单调栈。。。我。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
const int nmax=80005;
int a[nmax],s[nmax];
int main(){
	ll ans=0;
	int n;scanf("%d",&n);
	rep(i,1,n) scanf("%d",&a[i]);
	s[1]=1;ll r=1;
	rep(i,2,n){
		while(r&&a[s[r]]<=a[i]) r--;
		ans+=r;s[++r]=i;
	}
	printf("%lld\n",ans);
	return 0;
}

bzoj1657：

还是单调栈。。。我。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int x;char c;
int read(){
	x=0;c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=50005;
int h[nmax],v[nmax],s[nmax],ans[nmax];
int main(){
	int N=read();
	rep(i,1,N) h[i]=read(),v[i]=read();
	s[1]=1;;int r=1;
	rep(i,2,N){
		while(r&&h[s[r]]<h[i]) ans[i]+=v[s[r]],r--;
		s[++r]=i;
	}
	s[1]=N;r=1;
	dwn(i,N-1,1){
		while(r&&h[s[r]]<h[i]) ans[i]+=v[s[r]],r--;
		s[++r]=i;
	}
	int res=0;
	rep(i,1,N) res=max(res,ans[i]);
	printf("%d\n",res);
	return 0;
}

bzoj1655：

背包dp+高精度。。。看清楚数据范围看清楚数据范围看清楚数据范围！！！

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
int dp[1005][1005];
void add(int x,int y){
	rep(i,1,dp[y][0]) dp[x][i]+=dp[y][i];
	dp[x][0]=max(dp[x][0],dp[y][0]);
	rep(i,1,dp[x][0])
	  if(dp[x][i]>=10) dp[x][i]-=10,dp[x][i+1]++;
	if(dp[x][dp[x][0]+1]) dp[x][0]++;
}
int main(){
	int n,k;scanf("%d%d",&n,&k);
	dp[0][0]=dp[0][1]=1;
	rep(i,1,k) rep(j,i,n) add(j,j-i);
	dwn(i,dp[n][0],1) printf("%d",dp[n][i]);
	return 0;
}

bzoj1654：

tarjan即可

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=10005;
const int maxn=50005;
const int inf=0x7f7f7f7f;
struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
}
int sccno[nmax],pre[nmax],ans=0,dfs_clock=0,scc_cnt=0;
stack<int>s;
int dfs(int x){
	int lowu=pre[x]=++dfs_clock;s.push(x);
	qwq(x) {
		int to=o->to;
		if(!pre[to]) lowu=min(lowu,dfs(to));
		else if(!sccno[to]) lowu=min(lowu,pre[to]);
	}
	if(lowu==pre[x]){
		int sum=0;scc_cnt++;
		while(1){
			sum++;
			int X=s.top();s.pop();
			sccno[X]=scc_cnt;
			if(x==X) break;
		}
		if(sum!=1) ans++;
	}
	return lowu;
}
int main(){
	int N=read(),M=read(),u,v;
	rep(i,1,M) u=read(),v=read(),add(u,v);
	rep(i,1,N) if(!pre[i]) dfs(i);
	printf("%d\n",ans);
	return 0;
}

bzoj1653：

N很小，求出所有全排列后利用杨辉三角形计算次数即可

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int a[11],dp[11][11];
int main(){
	int n,m,tmp;scanf("%d%d",&n,&m);
	rep(i,1,n) a[i]=i;
	dp[1][1]=1;
	rep(i,2,n) rep(j,1,i) dp[i][j]=dp[i-1][j]+dp[i-1][j-1];
	while(1){
		tmp=0;
		rep(i,1,n) tmp+=a[i]*dp[n][i];
		if(tmp==m) break;
		next_permutation(a+1,a+n+1);
	}
	printf("%d",a[1]);
	rep(i,2,n) printf(" %d",a[i]);
	printf("\n");
	return 0;
}

bzoj1652：

和做过的差不多，可以压掉一维不会空间没炸就没压了。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=2005;
int a[nmax],f[nmax][nmax];
int main(){
	int n;scanf("%d",&n);
	rep(i,1,n) scanf("%d",&a[i]);
	rep(i,1,n) {
		for(int j=1;j+i-1<=n;j++){
			f[j][j+i-1]=max(f[j+1][j+i-1]+(n-i+1)*a[j],f[j][j+i-2]+(n-i+1)*a[j+i-1]);
		}
	}
	printf("%d\n",f[1][n]);
	return 0;
}

bzoj1651：

差分序列即可

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=1000005;
int a[nmax];
int main(){
	int N=read(),u,v,tot=-1,ans=0;
	rep(i,1,N) u=read(),v=read(),a[u]+=1,a[v+1]-=1,tot=max(tot,v);
	rep(i,1,tot) ans=max(ans,a[i]+=a[i-1]);
	printf("%d\n",ans);
	return 0;
}

❤bzoj1649：

跟背包dp差不多

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x));
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=1005;
int dp[nmax][nmax];
struct node{
	int s,t,w,c;
	bool operator<(const node&rhs)const{
	  return s<rhs.s||s==rhs.s&&t<rhs.t;}
};
node nodes[nmax*10];
int main(){
	int L=read(),N=read(),B=read();
	rep(i,1,N) {
		node &o=nodes[i];
		o.s=o.t=read(),o.t+=read(),o.w=read(),o.c=read();
	}
	sort(nodes+1,nodes+N+1);
	clr(dp,-1);dp[0][0]=0;
	rep(i,1,N) dwn(j,B,nodes[i].c)
	  if(dp[nodes[i].s][j-nodes[i].c]!=-1)
	    dp[nodes[i].t][j]=max(dp[nodes[i].s][j-nodes[i].c]+nodes[i].w,dp[nodes[i].t][j]);
	int ans=-1;
	rep(i,1,B) if(dp[L][i]!=-1) ans=max(ans,dp[L][i]);
	printf("%d\n",ans);
	return 0;
}