8.3-8.7 usaco

summary：38

vijos1002：青蛙跳河。

dp+压缩。距离大于100可以直接%100.然后数据范围小了很多可以dp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 f[100000];
int a[105];
bool vis[100000];
int main(){
	int l,s,t,m;
	scanf("%d%d%d%d",&l,&s,&t,&m);
	rep(i,1,m) scanf("%d",&a[i]);
	sort(a+1,a+m+1);
	if(s==t){
		int ans=0;
		rep(i,1,m) if(!(a[i]%s)) ans++;
		printf("%d\n",ans);return 0;
	}
	int tmp,cnt=0;
	rep(i,1,m){
		if((tmp=a[i]-a[i-1])>100) vis[cnt+=100]=1;
		else vis[cnt+=tmp]=1;
	}
	rep(i,0,cnt+t) f[i]=100;f[0]=0;
	rep(i,0,cnt+t) rep(j,s,t) if(i>=j) f[i]=min(f[i],f[i-j]+vis[i]);
	int ans=105;
	rep(i,cnt,cnt+t) ans=min(ans,f[i]);
	printf("%d\n",ans);
	return 0;
}

vijos1843：货车运输

最大生成树+lca。然而链剖太久没打WA了很久。id和idx容易混淆注意。是最大生成树森林处理一下，将边的权值弄到点的权值上。

#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)
#define lson x<<1,l,m
#define rson x<<1|1,m+1,r
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 dist,from,to;edge *next;
	bool operator<(const edge &rhs)const{
	  return dist>rhs.dist;}
};
edge e[maxn],edges[maxn<<1],*pt=edges,*head[nmax];
int fa[nmax],n,m,pre[nmax],son[nmax],size[nmax],dep[nmax],tp[nmax],id[nmax],idx[nmax],w[nmax],sum[nmax<<2];
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 find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
void dfs(int x){
	size[x]=1;
	qwq(x) if(o->to!=pre[x]){
		int to=o->to;pre[to]=x;dep[to]=dep[x]+1;w[to]=o->dist;
		dfs(to);size[x]+=size[to];
		if(!son[x]||size[to]>size[son[x]]) son[x]=to;
	}
}
void DFS(int x,int top){
	id[++id[0]]=x;idx[x]=id[0];tp[x]=top;
	if(son[x]) DFS(son[x],top);
	qwq(x) if(!idx[o->to]) DFS(o->to,o->to);
}
void build(int x,int l,int r){
	if(l==r) {
		sum[x]=w[id[l]];return ;
	}
	int m=(l+r)>>1;build(lson);build(rson);sum[x]=min(sum[x<<1],sum[x<<1|1]);
}
int query(int tl,int tr,int x,int l,int r){
	if(tl<=l&&tr>=r) return sum[x];
	int m=(l+r)>>1;int ans=inf;
	if(tl<=m) ans=min(ans,query(tl,tr,lson));
	if(tr>m) ans=min(ans,query(tl,tr,rson));
	return ans;
}
int qmax(int a,int b){
	if(find(a)!=find(b)) return -1;
	int ans=inf;
	while(tp[a]!=tp[b]){
		if(dep[tp[a]]>dep[tp[b]]) swap(a,b);
		ans=min(ans,query(idx[tp[b]],idx[b],1,1,n));
		b=pre[tp[b]];
	}
	if(dep[a]>dep[b]) swap(a,b);
	if(idx[a]==idx[b]) return ans;
	ans=min(ans,query(idx[a]+1,idx[b],1,1,n));
	return ans;
}
int main(){
	n=read(),m=read();
	rep(i,1,m) {
		edge &o=e[i];
		o.from=read(),o.to=read(),o.dist=read();
	}

	sort(e+1,e+m+1);
	rep(i,1,n) fa[i]=i;
	rep(i,1,m){
		edge &o=e[i];
		int ta=find(o.from),tb=find(o.to);
		if(ta!=tb){
			add(o.from,o.to,o.dist);
			fa[ta]=tb;
		}
	}

	clr(son,0);clr(idx,0);id[0]=0;
	rep(i,1,n) if(!pre[i]) dep[i]=0,dfs(i);
	rep(i,1,n) if(!idx[i]) DFS(i,i);

	build(1,1,n);
	int Q=read(),u,v;
	rep(i,1,Q){
		u=read(),v=read();
		printf("%d\n",qmax(u,v));
	}
}

bzoj1650：

二分答案+贪心判断。和noip2015day2t1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 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 a[nmax],N,M,L;
bool check(int x){
	int ans=0,pre=0;
	rep(i,1,N) {
		if(a[i]-a[pre]<x) ans++;
		else pre=i;
	}
	if(ans<=M) return true;
	return false;
}
int main(){
	L=read(),N=read(),M=read();
	rep(i,1,N) a[i]=read();
	sort(a+1,a+N+1);
	int l=0,r=L,mid,ans=0;
	while(l<=r){
		mid=(l+r)>>1;
		if(check(mid)) ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",ans);
	return 0;
}

bzoj1648：

边数点数少直接暴力dfs

#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 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=10005;
const int inf=0x7f7f7f7f;
struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
int a[nmax],ans[nmax];
bool vis[nmax];
void add(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
}
void dfs(int x){
	vis[x]=true;ans[x]++;
	qwq(x) if(!vis[o->to]) dfs(o->to);
}
int main(){
	int K=read(),N=read(),M=read(),u,v;
	rep(i,1,K) a[i]=read();
	rep(i,1,M) u=read(),v=read(),add(u,v);
	rep(i,1,K) clr(vis,false),dfs(a[i]);
	int res=0;
	rep(i,1,N) if(ans[i]==K) res++;
	printf("%d\n",res);
	return 0;
}

bzoj1646：

简单bfs；

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=200005;
int q[nmax],dist[nmax];
bool vis[nmax];
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	int l=1,r=1;
	clr(vis,false);vis[n]=true;
	dist[n]=0;
	q[r]=n;
	while(l<=r){
		int x=q[l];l++;
		if(x==k){
			printf("%d\n",dist[x]);return 0;
		}
		if(x-1>=0&&!vis[x-1]) q[++r]=x-1,vis[x-1]=true,dist[x-1]=dist[x]+1;
		if(x+1<=200000&&!vis[x+1]) q[++r]=x+1,vis[x+1]=true,dist[x+1]=dist[x]+1;
		if(x+x<=200000&&!vis[x+x]) q[++r]=x+x,vis[x+x]=true,dist[x+x]=dist[x]+1;
	}
	return 0;
}

bzoj1644：

bfs。不知道为什么我这样子写WA了。。。放一下代码求指点。。挖坑有空再写。

#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))
char s[105];
bool map[105][105];
int f[105][105],from[105][105];
queue<int>q,Q;
int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};
int main(){
	int n,sa,sb,ta,tb;scanf("%d",&n);
	rep(i,1,n){
		scanf("%s",s+1);
		rep(j,1,n) {
			map[i][j]=s[j]=='x'?0:1;
		    if(s[j]=='A') sa=i,sb=j;
		    else if(s[j]=='B') ta=i,tb=j;
		}
	}
	clr(f,0x7f);
	f[sa][sb]=0;
	q.push(sa);Q.push(sb);
	int x,y,tx,ty;
	while(!q.empty()){
		x=q.front(),y=Q.front();q.pop();Q.pop();
		rep(i,1,4){
			tx=x+xx[i],ty=y+yy[i];
			if(tx<=0||ty<=0||tx>n||ty>n||!map[tx][ty]) continue;
			if(from[x][y]==i||x==sa&&y==sb) {
				if(f[x][y]<f[tx][ty])
			      q.push(tx),Q.push(ty),f[tx][ty]=f[x][y],from[tx][ty]=i;
			}
			else  if(f[x][y]<f[tx][ty]-1)
			   q.push(tx),Q.push(ty),f[tx][ty]=f[x][y]+1,from[tx][ty]=i;
		}
	}
	printf("%d\n",f[ta][tb]);
	return 0;
}
/*
5
..Bx.
.xxA.
...x.
.x...
..x..*/

bzoj1643：

O（n^3）暴力，然后第四个数可以推出来。。因为写过一道搜索+剪枝有一个剪枝类似这样子所以会。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
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[105];
int main(){
	int n,u,temp,orz,ans=0;
	scanf("%d",&n);
	int tmp=(int)sqrt(n)+1;
	rep(i,0,tmp) a[i]=i*i;
	rep(i,0,tmp) {
		rep(j,0,tmp) if(a[i]+a[j]<=n){
			rep(k,0,tmp) if((u=a[i]+a[j]+a[k])<=n){
				temp=n-u,orz=sqrt(temp);
				if(orz*orz==temp) ans++;
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}

bzoj1641：

floyed即可。。。然而一看到题目就二话不说二分答案+暴力dfs（总是觉得应该可以卡时过去。然而还是tle了。。、不过这种做法应该是对的吧

/*#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=305;
const int maxn=25005;
const int inf=0x7f7f7f7f;
struct edge{
	int to,dist;edge *next;
	bool flag;
};
edge edges[maxn],*pt=edges,*head[nmax];
int N,M,Q;
bool vis[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
}
bool dfs(int x,int t){
	vis[x]=true;
	if(x==t) return true;
	qwq(x) if(!vis[o->to]&&o->flag){
		if(dfs(o->to,t)) return true;
	}
	return false;
}
bool check(int x,int s,int t){
	rep(i,1,N) qwq(i) if(o->dist<=x) o->flag=true;else o->flag=false;
	clr(vis,false);
	if(dfs(s,t)) return true;
	return false;
}
int main(){
	N=read(),M=read(),Q=read();
	int u,v,d,sum=0;
	rep(i,1,M) u=read(),v=read(),d=read(),add(u,v,d),sum=max(sum,d);
	rep(i,1,Q) {
		u=read(),v=read();
		int l=0,r=sum,ans=-1,mid;
	    while(l<=r){
		   mid=(l+r)>>1;
		   if(check(mid,u,v)) ans=mid,r=mid-1;
		   else l=mid+1;
	   }
	   printf("%d\n",ans);
	}
}*/
#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=305;
const int maxn=25005;
const int inf=0x7f7f7f7f;
int dist[nmax][nmax];
int main(){
	int N=read(),M=read(),Q=read(),u,v,d;
	clr(dist,0x7f);
	rep(i,1,M) u=read(),v=read(),d=read(),dist[u][v]=d;
	rep(k,1,N) rep(i,1,N) rep(j,1,N) if(dist[i][k]!=inf&&dist[k][j]!=inf)
	  dist[i][j]=min(dist[i][j],max(dist[i][k],dist[k][j]));
	rep(i,1,Q) {
		u=read(),v=read();
		if(dist[u][v]==inf) printf("-1\n");
		else printf("%d\n",dist[u][v]);
	}
	return 0;
}

bzoj1640：

直接模拟。。。然而没有看清题意WA了唉。。。

#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],ans[nmax];
char c[5];
int main(){
	int n;scanf("%d",&n);
	rep(i,1,n){
		scanf("%s",c);
		a[i]=c[0]-'A';
	}
	int l=1,r=n,tmp;
	rep(k,1,n){
		if(a[l]<a[r]) ans[k]=a[l],l++;
		else if(a[l]>a[r]) ans[k]=a[r],r--;
		else{
			for(int i=l,j=r;i<=j;i++,j--){
				if(a[i]<a[j]){
					tmp=l;break;
				}else if(a[i]>a[j]){
					tmp=r;break;
				}else if(i==j){
					tmp=l;break;
				}
			}
			ans[k]=a[tmp];
			tmp==l?l++:r--;
		}
	}
	rep(i,1,n){
		putchar('A'+ans[i]);
		if(i%80==0) putchar('\n');
	}
	return 0;
}
/*
10
A
C
D
E
F
B
E
D
C
B
*/

bzoj1639：

二分答案+贪心判断。这种模拟总有感觉自己会写WA，怀疑自己写错。但是只要用模拟一下数据的运算过程还是很容易判断写法的对错的。这也是模拟的调法吧。

#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=100005;
const int inf=0x7f7f7f7f;
int N,M,a[nmax];
bool check(int x){
	int ans=0,cur=1,tmp;
	while(1){
		tmp=a[cur];
		if(tmp>x) return false;
		while(tmp<=x&&cur<=N) cur++,tmp+=a[cur];
		ans++;
		if(cur>N) break;
	}
	if(ans<=M) return true;
	return false;
}
int main(){
	N=read(),M=read();
	rep(i,1,N) a[i]=read();
	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;
}

bzoj1638：

dfs。不会做，orz了hzwer（我原本的做法求方案数又是抱着应该可以卡过去的想法暴力dfs。。。stopstopstop小搜怡情爆搜伤身啊。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define 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)
#define qaq(x) for(edge *o=h[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=5005;
const int maxn=50005;
const int inf=0x7f7f7f7f;
bool in[nmax];
int s[maxn],t[maxn];
int f[nmax],dp[nmax];
struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
edge e[maxn],*p=e,*h[nmax];
void add(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v){
	p->to=v;p->next=h[u];h[u]=p++;
}
void dfs(int x){
	if(!head[x]) {
		dp[x]=1;return ;
	}
	qwq(x){
		if(!dp[o->to]) dfs(o->to);
		dp[x]+=dp[o->to];
	}
}
void DFS(int x){
	if(!h[x]) {
		f[x]=1;return ;
	}
	qaq(x){
		if(!f[o->to]) DFS(o->to);
		f[x]+=f[o->to];
	}
}
int main(){
	int N=read(),M=read(),u,v;
	rep(i,1,M) u=read(),v=read(),add(u,v),adde(v,u),in[v]=true,s[i]=u,t[i]=v;
	rep(i,1,N) if(!in[i]) dfs(i);
	DFS(N);
	int ans=0;
	rep(i,1,M) ans=max(ans,dp[t[i]]*f[s[i]]);
	printf("%d\n",ans);
	return 0;
}

bzoj1637：

1就+1,0就-1，然后找乱搞就可以了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define 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;
struct node{
	int cur,x;
	bool operator<(const node&rhs)const{
	  return cur<rhs.cur;}
};
node nodes[nmax];
int sum[nmax],ll[nmax+nmax],rr[nmax+nmax];
int main(){
	int n=read();
	rep(i,1,n) nodes[i].x=read(),nodes[i].cur=read();
	sort(nodes+1,nodes+n+1);
	rep(i,1,n) {
		node&o=nodes[i];
		if(o.x) sum[i]=sum[i-1]+1;
		else sum[i]=sum[i-1]-1;
		if(!ll[sum[i]+n]) ll[sum[i]+n]=i;
		else rr[sum[i]+n]=i;
	}
	int ans=0;
	rep(i,0,n+n){
		if(!rr[i]) continue;
		ans=max(ans,nodes[rr[i]].cur-nodes[ll[i]+1].cur);
	}
	printf("%d\n",ans);
	return 0;
}

bzoj1636：

多次求区间最大最小差值。ST和线段树都可以搞。没写过ST写一下还是挺好写的。

#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;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
char s[100];
void print(int x){
	int cur=0;
	if(!x) {
		putchar('0');putchar('\n');
		return;
	}
	while(x){
		s[++cur]=x%10+'0',x/=10;
	}
	dwn(i,cur,1) putchar(s[i]);
	putchar('\n');
}
const int nmax=50005;
int a[nmax],f[nmax][20],g[nmax][20];
int main(){
	int N=read(),Q=read(),u,v;
	rep(i,1,N) a[i]=read(),f[i][0]=g[i][0]=a[i];
	for(int k=1;(1<<k)<=N;k++)
	  for(int i=1;i+(1<<k)-1<=N;i++)
	    f[i][k]=max(f[i][k-1],f[i+(1<<k-1)][k-1]),g[i][k]=min(g[i][k-1],g[i+(1<<k-1)][k-1]);
	rep(i,1,Q){
		u=read(),v=read();
		int tmp;
		for(tmp=0;(1<<tmp)<=v-u+1;tmp++);tmp--;
		print(max(f[u][tmp],f[v-(1<<tmp)+1][tmp])-min(g[u][tmp],g[v-(1<<tmp)+1][tmp]));
	}
	return 0;
}

bzoj1635：

差分序列！！！感觉好高大上的东西。就是可以区间内加减一个数。然后就可以啦。

#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;
int ans[nmax];
struct node{
	int x,y;
	bool operator<(const node&rhs)const{
	  return x<rhs.x||x==rhs.x&&y<rhs.y;}
};
node nodes[nmax];
int main(){
	int N=read(),I=read(),H=read(),M=read(),u,v;
	rep(i,1,M) {
		node&o=nodes[i];
		o.x=read(),o.y=read();
		if(o.x>o.y) swap(o.x,o.y);
	}
	sort(nodes+1,nodes+M+1);
	rep(i,1,M){
		if(nodes[i].x==nodes[i-1].x&&nodes[i].y==nodes[i-1].y) continue;
		ans[nodes[i].x+1]--;ans[nodes[i].y]++;
	}
	rep(i,1,N) printf("%d\n",(ans[i]+=ans[i-1])+H);printf("\n");
	return 0;
}

bzoj1634：

贪心。对于AB判断A前B后或者A后B前哪一种更优可以推出表达式。那么久可以确定顺序了。以点破面？。。。

#include<cstdio>
#include<cstring>
#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
const int nmax=100005;
struct node{
	int w,t,num;
	bool operator<(const node&rhs)const{
	  return rhs.w*t>w*rhs.t;}
};
node nodes[nmax];
int main(){
	int N;scanf("%d",&N);
	rep(i,1,N) scanf("%d%d",&nodes[i].w,&nodes[i].t),nodes[i].num=i;
	sort(nodes+1,nodes+N+1);
	ll sum=0,ans=0;
	rep(i,1,N) ans+=sum*nodes[i].t<<1,sum+=nodes[i].w;
	printf("%lld\n",ans);
	return 0;
}

bzoj1632：

bfs。多加一维来乱搞（然而我一直调不出来。。。挖坑。啊我为什么bfs总是WA啊。。。

bzoj1631：

两遍spfa就可以了。。。将边反过来存储就又变成单源最短路了。。以前见过。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#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 qwq(x) for(edge *o=head[x];o;o=o->next)
#define qaq(x) for(edge *o=h[x];o;o=o->next)
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 f*x;
}
const int nmax=1005;
const int maxn=100005;
const int inf=0x7f7f7f7f;
struct edge{
	int to,dist;edge*next;
};
edge edges[maxn],*pt=edges,*head[nmax];
edge e[maxn],*p=e,*h[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v,int d){
	p->to=v;p->dist=d;p->next=h[u];h[u]=p++;
}
queue<int>q;
int dist[nmax],dis[nmax];
bool inq[nmax];
void spfa(int t){
	clr(inq,false);inq[t]=1;
	clr(dist,0x7f);dist[t]=0;
	q.push(t);
	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;
		}
	}
}
void SPFA(int t){
	clr(inq,false);inq[t]=1;
	clr(dis,0x7f);dis[t]=0;
	q.push(t);
	while(!q.empty()){
		int x=q.front();q.pop();inq[x]=false;
		qaq(x) if(dis[o->to]>dis[x]+o->dist){
			dis[o->to]=dis[x]+o->dist;
			if(!inq[o->to]) q.push(o->to),inq[o->to]=true;
		}
	}
}
int main(){
	int N=read(),M=read(),T=read(),u,v,d;
	rep(i,1,M){
		u=read(),v=read(),d=read();
		add(u,v,d);adde(v,u,d);
	}
	spfa(T);SPFA(T);
	int ans=-1;
	rep(i,1,N) ans=max(ans,dist[i]+dis[i]);
	printf("%d\n",ans);
	return 0;
}

bzoj1630：

这可是道背包dp神题啊。挖坑似乎可以前缀和优化+滚动数组。。。然而我只会滚动数组。（防止mle。好劲啊这是。。。

#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;
const int maxn=1e3+5;
const int mod=1e6;
int dp[nmax],cnt[maxn];
int main(){
	int T=read(),A=read(),S=read(),B=read(),u;
	rep(i,1,A) u=read(),cnt[u]++;
	dp[0]=1;
	rep(i,1,T) dwn(j,B,1) rep(k,1,cnt[i]) {
		if(j<k) break;
		dp[j]=(dp[j]+dp[j-k])%mod;
	}
	int ans=0;
	rep(i,S,B) ans=(ans+dp[i])%mod;
	printf("%d\n",ans);
	return 0;
}

bzoj1629：

贪心。。AB谁前谁后可以判断。和1634是同一种类型题。

#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=50005;
const int inf=0x7f7f7f7f;
struct node{
	int w,t;
	bool operator<(const node&rhs)const{
	  return max(-t,w-rhs.t)<max(-rhs.t,rhs.w-t);}
};
node nodes[nmax];
int main(){
	int n=read();
	rep(i,1,n) nodes[i].w=read(),nodes[i].t=read();
	sort(nodes+1,nodes+n+1);
	int ans=-inf,sum=0;
	rep(i,1,n) ans=max(ans,sum-nodes[i].t),sum+=nodes[i].w;
	printf("%d\n",ans);
	return 0;
}

bzoj1628：

单调栈！边界没有处理好！！！注意边界。。

#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=50005;
int a[nmax],s[nmax];
int main(){
	int N=read(),W=read(),u,v;
	rep(i,1,N) u=read(),a[i]=read();
	int r=0,ans=0;
	rep(i,1,N){
		while(r&&s[r]>a[i]) r--;
		if(s[r]!=a[i]) ans++;
		s[++r]=a[i];
	}
	printf("%d\n",ans);
	return 0;
}

bzoj1627：

简单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))
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 f*x;
}
const int nmax=1005;
const int orz=500;
bool on[nmax][nmax],vis[nmax][nmax];
int dist[nmax][nmax];
struct node{
	int x,y;
	node(int x,int y):x(x),y(y){}
};
queue<node>q;
int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};
int main(){
	int ta=read(),tb=read(),N=read(),u,v,d;
	ta+=orz,tb+=orz;
	clr(vis,false);clr(on,true);
	rep(i,1,N) u=read(),v=read(),on[u+orz][v+orz]=false;
	q.push(node(orz,orz));vis[orz][orz]=true;
	while(!q.empty()){
		node t=q.front();q.pop();
		rep(i,1,4){
			int tx=t.x+xx[i],ty=t.y+yy[i];
			if(tx&&ty&&tx<=1000&&ty<=1000&&!vis[tx][ty]&&on[tx][ty]) {
				vis[tx][ty]=true;dist[tx][ty]=dist[t.x][t.y]+1;
				q.push(node(tx,ty));
				if(tx==ta&&ty==tb){
					printf("%d\n",dist[tx][ty]);
					return 0;
				}
			}
		}
	}
	return 0;
}

bzoj1626：

最小生成树。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
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=1000005;
int fa[nmax],xi[nmax],yi[nmax];
struct edge{
	int from,to;double dist;
	bool operator<(const edge&rhs)const{
	  return dist<rhs.dist;}
};
edge edges[maxn];
int find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
int main(){
	int N=read(),M=read(),u,v,ta,tb;
	rep(i,1,N) xi[i]=read(),yi[i]=read();
	rep(i,1,N) fa[i]=i;
	rep(i,1,M){
		u=read(),v=read();
		ta=find(u),tb=find(v);
		if(ta!=tb) fa[ta]=tb;
	}
	int cnt=0;
	rep(i,1,N-1) rep(j,i+1,N) {
		edge &o=edges[++cnt];
		o.from=i,o.to=j;
		o.dist=sqrt((double)(xi[i]-xi[j])*(xi[i]-xi[j])+(double)(yi[i]-yi[j])*(yi[i]-yi[j]));
	}
	sort(edges+1,edges+cnt+1);
	//rep(i,1,cnt) printf("%d %d %lf\n",edges[i].from,edges[i].to,edges[i].dist);
	double ans=0;int res=M;
	rep(i,1,cnt){
		ta=find(edges[i].from);tb=find(edges[i].to);
		if(ta!=tb) fa[ta]=tb,ans+=edges[i].dist,res++;
		if(res==N-1) break;
	}
	printf("%.2lf\n",ans);
	return 0;
}

bzoj1625：

背包dp。。。

#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=3500;
const int maxn=13000;
int w[nmax],c[nmax];
int dp[maxn];
int main(){
	int N=read(),M=read();
	rep(i,1,N) w[i]=read(),c[i]=read();
	rep(i,1,N) dwn(j,M,w[i]) dp[j]=max(dp[j],dp[j-w[i]]+c[i]);
	printf("%d\n",dp[M]);
	return 0;
}

bzoj1624：

floyed。。。

#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=105;
const int maxn=10005;
int dist[nmax][nmax];
int ans[maxn];
int main(){
	int N=read(),M=read();
	rep(i,1,M) ans[i]=read();
	rep(i,1,N) rep(j,1,N) dist[i][j]=read();
	rep(k,1,N) rep(i,1,N) rep(j,1,N)
	  dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
	int res=0;
	rep(i,1,M-1) res+=dist[ans[i]][ans[i+1]];
	printf("%d\n",res);
	return 0;
}

bzoj1623：

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 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 a[nmax];
int main(){
	int N=read(),M=read(),D=read(),L=read();
	rep(i,1,N) a[i]=read();
	sort(a+1,a+N+1);
	int ans=0;
	rep(i,1,N){
		if(a[i]-ans/M*D>=L) ans++;
	}
	printf("%d\n",ans);
	return 0;
}

bzoj1622：

乱搞。。(这种模拟题我总是很虚啊。。。

#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))
const int nmax=1005;
const int maxn=105;
char s[nmax][nmax],t[maxn][35];
int ans[nmax];
int lena[nmax],lenb[nmax];
int main(){
	int N,M;
	scanf("%d%d",&N,&M);
	rep(i,1,N) {
		scanf("%s",s[i]);
		lena[i]=strlen(s[i]);
		rep(j,0,lena[i]-1)
		  if(s[i][j]<'a') s[i][j]+=32;
	}
	rep(i,1,M) {
		scanf("%s",t[i]);
		lenb[i]=strlen(t[i]);
		rep(j,0,lenb[i]-1)
		  if(t[i][j]<'a') t[i][j]+=32;
	}
	rep(i,1,N) rep(j,1,M) {
		int p=0,pt=0;
		while(p<lena[i]&&pt<lenb[j]){
			if(s[i][p]==t[j][pt]) pt++;
			p++;
		}
		if(pt==lenb[j]) ans[i]++;
	}
	rep(i,1,N) printf("%d\n",ans[i]);
	return 0;
}

bzoj1621：

递归。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
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;
}
int n,K;
int dfs(int x){
	if((x+K)%2) return 1;
	if(x<=K) return 1;
	return dfs((x+K)/2)+dfs((x-K)/2);
}
int main(){
	n=read(),K=read();
	printf("%d\n",dfs(n));
	return 0;
}

bzoj1620：

二分答案+贪心判断。。。

#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=100005;
const int inf=0x7f7f7f7f;
int N,M,a[nmax];
bool check(int x){
	int ans=0,cur=1,tmp;
	while(1){
		tmp=a[cur];
		if(tmp>x) return false;
		while(tmp<=x&&cur<=N) cur++,tmp+=a[cur];
		ans++;
		if(cur>N) break;
	}
	if(ans<=M) return true;
	return false;
}
int main(){
	N=read(),M=read();
	rep(i,1,N) a[i]=read();
	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;
}

bzoj1618：

背包dp。。。

#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=60005;
const int maxn=105;
const int inf=0x7f7f7f7f;
int w[maxn],c[maxn];
int f[nmax];
int main(){
	int n=read(),h=read();
	rep(i,1,n) w[i]=read(),c[i]=read();
	clr(f,0x7f);f[0]=0;
	rep(i,1,n) rep(j,0,h) if(f[j]!=inf) f[j+w[i]]=min(f[j+w[i]],f[j]+c[i]);
	int ans=inf;
	rep(i,h,h+5005) ans=min(ans,f[i]);
	printf("%d\n",ans);
	return 0;
}

bzoj1617：

区间dp。。。

#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;
int read(){
	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=2505;
const int inf=0x7f7f7f7f;
int cost[nmax],dp[nmax];
int main(){
	int n=read(),m=read();
	rep(i,1,n) cost[i]=read(),cost[i]+=cost[i-1];
	clr(dp,0x7f);dp[0]=0;
	rep(i,1,n) rep(j,0,i-1) dp[i]=min(dp[i],dp[j]+cost[i-j]+m+m);
	printf("%d\n",dp[n]-m);
	return 0;
}

bzoj1616：

多加一维然后dp转移即可。。。

#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=105;
const int maxn=20;
int dp[maxn][nmax][nmax],a[nmax][nmax];
char s[nmax];
int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};
int main(){
	int n=read(),m=read(),T=read();
	rep(i,1,n){
		scanf("%s",s);
		rep(j,1,m) a[i][j]=s[j-1]=='.'?1:0;
	}
	int sa=read(),ta=read(),sb=read(),tb=read();
	dp[0][sa][ta]=1;
	rep(i,1,T) rep(j,1,n) rep(k,1,m) {
		if(a[j][k]) {
			rep(o,1,4){
				int tx=j+xx[o],ty=k+yy[o];
				if(tx&&ty&&tx<=n&&ty<=m&&a[tx][ty]) dp[i][j][k]+=dp[i-1][tx][ty];
			}
		}else dp[i][j][k]=0;
	}
	printf("%d\n",dp[T][sb][tb]);
	return 0;
}

bzoj1615：

乱搞。。。

#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();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=205;
int x[nmax],y[nmax];
double ans[nmax];
int main(){
	int n=read();
	rep(i,1,n) x[i]=read(),y[i]=read();
	int cnt=0;bool f=false;
	rep(i,1,n) rep(j,1,n) if(i!=j){
		if(x[i]==x[j]) f=true;
		else ans[++cnt]=(double)(y[i]-y[j])/(x[i]-x[j]);
	}
	sort(ans+1,ans+cnt+1);
	int res=0;
	if(cnt) res++;
	rep(i,2,cnt) if(ans[i]!=ans[i-1]) res++;
	if(f) res++;
	printf("%d\n",res);
	return 0;
}

bzoj1612：

spfa求路径最小值。。。都是差不多的啦。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#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 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=1005;
const int maxn=10005;
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,k;
	node(int x,int k):x(x),k(k){}
};
queue<node>q;
int dist[nmax][nmax];
bool inq[nmax][nmax];
void spfa(int K,int n){
	clr(dist,0x7f);dist[1][0]=0;
	clr(inq,false);inq[1][0]=true;
	q.push(node(1,0));
	while(!q.empty()){
		node oo=q.front();q.pop();inq[oo.x][oo.k]=false;
		qwq(oo.x){
			int to=o->to;
			if(dist[to][oo.k]>max(dist[oo.x][oo.k],o->dist)) {
				dist[to][oo.k]=max(dist[oo.x][oo.k],o->dist);
				if(!inq[to][oo.k]){
					q.push(node(to,oo.k));inq[to][oo.k]=true;
				}
			}
			if(oo.k>=K) continue;
			if(dist[to][oo.k+1]>dist[oo.x][oo.k]){
				dist[to][oo.k+1]=dist[oo.x][oo.k];
				if(!inq[to][oo.k+1]){
					q.push(node(to,oo.k+1));inq[to][oo.k+1]=true;
				}
			}
		}
	}
	if(dist[n][K]==inf) printf("-1\n");
	else printf("%d\n",dist[n][K]);
	return ;
}
int main(){
	int n=read(),m=read(),p=read(),u,v,d;
	rep(i,1,m) u=read(),v=read(),d=read(),add(u,v,d);
	spfa(p,n);
	return 0;
}

bzoj1612：

暴力dfsQAQ。。。

#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 qwq(x) for(edge *o=head[x];o;o=o->next)
#define qaq(x) for(edge *o=h[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=105;
const int maxn=4505;
const int inf=0x7f7f7f7f;
struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
edge e[maxn],*p=e,*h[nmax];
int ans[nmax];
bool vis[nmax];
void add(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
	p->to=u;p->next=h[v];h[v]=p++;
}
void dfs(int x){
	qwq(x) if(!vis[o->to]) ans[o->to]++,vis[o->to]=true,dfs(o->to);
}
void DFS(int x){
	qaq(x) if(!vis[o->to]) ans[o->to]++,vis[o->to]=true,DFS(o->to);
}
int main(){
	int n=read(),m=read(),u,v;
	rep(i,1,m) u=read(),v=read(),add(u,v);
	rep(i,1,n) {
		clr(vis,false);dfs(i);
	}
	rep(i,1,n){
		clr(vis,false);DFS(i);
	}
	int res=0;
	rep(i,1,n) if(ans[i]==n-1) res++;
	printf("%d\n",res);
	return 0;
}

bzoj1611：

对于每个点，能转移到该点必定没有被炸，那么可以利用这一点性质在bfs时转移！。。

#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();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=305;
const int inf=0x7f7f7f7f;
int map[nmax][nmax],dist[nmax][nmax];
bool vis[nmax][nmax];
struct node{
	int x,y;
	node(int x,int y):x(x),y(y){}
};
queue<node>q;
int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};
int main(){
	int m=read(),u,v,d;
	clr(map,0x7f);
	rep(i,1,m) {
		u=read(),v=read(),d=read();map[u][v]=min(map[u][v],d);
		rep(k,1,4) if(u+xx[k]>=0&&v+yy[k]>=0) map[u+xx[k]][v+yy[k]]=min(d,map[u+xx[k]][v+yy[k]]);
	}
	if(map[0][0]==inf){
		printf("0\n");return 0;
	}
	q.push(node(0,0));
	clr(vis,false);vis[0][0]=true;dist[0][0]=0;
	while(!q.empty()){
		node o=q.front();q.pop();
		rep(i,1,4){
			int x=o.x,y=o.y,tx=x+xx[i],ty=y+yy[i];
			if(tx<0||ty<0||map[tx][ty]-1<=dist[x][y]||vis[tx][ty]) continue;
			if(map[tx][ty]==inf){
				printf("%d\n",dist[x][y]+1);return 0;
			}
			q.push(node(tx,ty));dist[tx][ty]=dist[x][y]+1;vis[tx][ty]=true;
		}
	}
	printf("-1\n");
	return 0;
}

bzoj4395：

暴力+bfs。多次bfs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 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=105;
const int maxn=20005;
const int inf=0x7f7f7f7f;
struct edge{
	int to,too;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax][nmax];
bool on[nmax][nmax],vis[nmax][nmax];
void add(int x,int y,int d,int w){
	pt->to=d;pt->too=w;pt->next=head[x][y];head[x][y]=pt++;
}

struct node{
	int a,b;
	node(int a,int b):a(a),b(b){}
};
queue<node>q;

int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};

int main(){
	int n=read(),m=read();
	rep(i,1,m){
		int u=read(),v=read(),d=read(),w=read();
		add(u,v,d,w);
	}
	int ans=1,last=1;on[1][1]=true;
	while(1){
	  q.push(node(1,1));
	  clr(vis,0);vis[1][1]=true;
	  while(!q.empty()){
		node x=q.front();q.pop();
		for(edge *o=head[x.a][x.b];o;o=o->next){
			if(!on[o->to][o->too]) on[o->to][o->too]=true,ans++;
		}
		rep(i,1,4){
			int tx=x.a+xx[i],ty=x.b+yy[i];
			if(tx&&ty&&tx<=n&&ty<=n){
				if(on[tx][ty]&&!vis[tx][ty])
				  vis[tx][ty]=true,q.push(node(tx,ty));
			}
		}
	  }
	  if(last==ans) break;
	  last=ans;
	}
	printf("%d\n",ans);
	return 0;
}

bzoj1318：

神题。。这种题都不知道怎么想出来的。。。orzhzwer后勉勉强强懂了。。。然而太神了这。。。

#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=1000015;
const int inf=0x7f7f7f7f;
int n,ans=-1,last[nmax],next[nmax],a[nmax];
ll sum[nmax];
bool vis[nmax];
void solve(int x){
	int len=0,rmax=inf;
	rep(i,x+1,n) {
		if(a[i]==1) break;
		if(!vis[a[i]]) vis[a[i]]=true;
		else {
			rmax=i;break;
		}
	}
	dwn(i,x,1){
		if(a[i]==1&&i!=x) break;
		rmax=min(rmax,next[i]);
		len=max(len,a[i]);
		if(i+len-1<=n&&i+len-1<rmax&&sum[i+len-1]-sum[i-1]==(ll)len*(len+1)/2) ans=max(ans,len);
	}
	rep(i,x+1,n) {
		if(a[i]==1) break;
		vis[a[i]]=false;
	}
}
void work(){
	clr(last,0x7f);
	dwn(i,n,1) {
		next[i]=last[a[i]];
		last[a[i]]=i;
	}
	rep(i,1,n) if(a[i]==1) solve(i);
}
int main(){
	n=read();
	rep(i,1,n) a[i]=read(),sum[i]=sum[i-1]+a[i];
	work();
	reverse(a+1,a+n+1);
	clr(sum,0);
	rep(i,1,n) sum[i]=sum[i-1]+a[i];
	work();
	printf("%d\n",ans);
	return 0;
}

bzoj1306：

搜索+剪枝！。我写了三个剪枝但是tle了。到网上一看然后再加一个剪枝就卡过去了。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(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;
}
int f[4]={3,1,0,0};
int ans[9],a[9],res=0,n;
void dfs(int x,int y){
	if(ans[x]>a[x]) return ;
	if(ans[x]+(n-y+1)*3<a[x]) return ;
	if(x==n) {
		res++;return ;
	}
	if(y==n){
		int tmp=a[x]-ans[x];
		if(tmp==2) return ;
		ans[y]+=f[tmp];
		dfs(x+1,x+2);
		ans[y]-=f[tmp];
	}else{
		ans[x]+=3;dfs(x,y+1);ans[x]-=3;
		ans[x]++,ans[y]++,dfs(x,y+1),ans[x]--,ans[y]--;
		ans[y]+=3;dfs(x,y+1),ans[y]-=3;
	}
}
int main(){
	n=read();
	rep(i,1,n) a[i]=read();
	dfs(1,2);
	printf("%d\n",res);
	return 0;
}

bzoj1303：

大于的就+1，小于的就-1，求中位数子串有多少个。。然后乱搞就可以了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define 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=200005;
const int inf=0x7f7f7f7f;
int a[nmax],cnt[nmax],sum[nmax];//qi ou
int main(){
	int n=read(),m=read(),cur,u;
	rep(i,1,n){
		a[i]=read();
		if(a[i]==m) cur=i;
		a[i]+=100000;
	}
	m+=100000;
	sum[100000]=1;a[0]=100000;
	rep(i,1,cur-1){
		if(a[i]>m) a[i]=a[i-1]+1;
		else a[i]=a[i-1]-1;
		if(i%2) cnt[a[i]]++;
		else sum[a[i]]++;
	}
	int ans=0;
	rep(i,cur,n){
		if(a[i]>m) a[i]=a[i-1]+1;
		else if(a[i]==m) a[i]=a[i-1];
		else a[i]=a[i-1]-1;
		if(i%2) ans+=sum[a[i]];
		else ans+=cnt[a[i]];
	}
	printf("%d\n",ans);
	return 0;
}