9.15 DP合集水表

9.15 DP合集水表

显然难了一些啊。

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凸多边形的三角剖分

瞄了一眼题解。

和蛤蛤的烦恼一样，裸的区间dp。

设f[i][j]表示i~j的点三角剖分最小代价。

显然\(f[i][i+1]=0,f[i][i+2]=w[i]*w[i+1]*w[i+2]\)

然后枚举i,j和哪个点剖。

\[f[l][r]=min(f[l][r],f[l][i]+f[i][r]+w[l]*w[r]*w[i]) (l<i<r)
\]

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "division"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0;rg char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x;
}
ll w[101],f[101][101];
int main(){
    #ifdef xzz
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
    #endif
    int n=gi(),r;
    rep(i,1,n)w[i]=w[i+n]=gi();
    n*=2;
    rep(i,3,n)f[i-2][i]=w[i-2]*w[i-1]*w[i];
    rep(siz,4,n/2)drep(l,n-siz+1,1){
	r=l+siz-1;
	f[l][r]=1e18;
	rep(i,l+1,r-1)f[l][r]=min(f[l][r],f[l][i]+f[i][r]+w[l]*w[r]*w[i]);
    }
    n/=2;
    ll ans=2e18;
    rep(i,1,n)ans=min(ans,f[i][i+n-1]);
    printf("%lld\n",ans);
    return 0;
}

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求最长公共子序列

字符序列的子序列是指从给定字符序列中随意地（不一定连续）去掉若干个字符（可能一个也不去掉）后所形成的字符序列。

SBT，切掉

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#include<cstring>
#define Fname "lcs"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0;rg char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x;
}
char a[5002],b[5002];
int f[5002][5002];
int main(){
    #ifdef xzz
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
    #endif
    scanf("%s%s",a+1,b+1);
    int lena=strlen(a+1)-1,lenb=strlen(b+1)-1;
    rep(i,1,lena)rep(j,1,lenb)
	if(a[i]==b[j])f[i][j]=f[i-1][j-1]+1;
	else f[i][j]=max(f[i-1][j],f[i][j-1]);
    printf("%d\n",f[lena][lenb]);
    return 0;
}

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卡车更新问题

设f[i][j]表示第i天，上次更新是j就好了

主要难在输出方案，恶心细节多

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "truck"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0;rg char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x;
}
#define db double
db R[23],U[23],C[23];
db f[23][23];
int g[23][23];
il vd Min(db&a,int&b,db c,int d){if(c>a)a=c,b=d;}
//f[i][j] 到第i天，上次更新是第j天的总回收额max
bool yes[23];
int main(){
    #ifdef xzz
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
    #endif
    int n=gi(),k=gi();
    rep(i,0,k)scanf("%lf",&R[i]);
    rep(i,0,k)scanf("%lf",&U[i]);
    rep(i,0,k)scanf("%lf",&C[i]);
    rep(i,1,n+1)rep(j,0,k)f[i][j]=-233333.33;
    f[1][0]=R[0]-U[0];
    rep(i,1,n-1)rep(j,0,i-1){
	if(i-j<k)Min(f[i+1][j],g[i+1][j],f[i][j]+R[i-j]-U[i-j],j);
	Min(f[i+1][i],g[i+1][i],f[i][j]+R[0]-U[0]-C[i-j],j);
    }
    int ans=n-1;
    rep(i,n-k,n-2)if(f[n][i]>f[n][ans])ans=i;
    printf("%.1f\n",f[n][ans]);
    int x=n,y=ans;
    while(x){
	if(y==x-1)yes[x]=1;
        y=g[x][y],--x;
    }x=n,y=ans;
    printf("%d 0 %.1f\n",1,R[0]-U[0]);
    y=1;
    rep(i,2,n){
	if(yes[i])printf("%d 1 %.1f\n",i,R[0]-U[0]-C[y]),y=1;
	else printf("%d 0 %.1f\n",i,R[y]-U[y]),++y;
    }
    return 0;
}

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选课

裸的树形背包但要输出方案。GG

不会

为了挽尊，特写spj一个。。。

#include<bits/stdc++.h>
using namespace std;
int main(int argc,char *argv[])
{
    FILE *in,*out,*ans,*score,*report;
    out=fopen(argv[2],"r");
    ans=fopen(argv[3],"r");
    score=fopen(argv[5],"w");
    report=fopen(argv[6],"w");
    int a=-2333,b=-6666;
    fscanf(out,"%d",&a);
    fscanf(ans,"%d",&b);
    if(a==b){
	while(fscanf(out,"%d",&a)==1||fscanf(ans,"%d",&b)==1)
	    if(a!=b){fprintf(score,"3"),fprintf(report,"方案错了。。。3分");return 0;}
	fprintf(score,"10"),fprintf(report,"泥太有才了！");
    }
    else fprintf(score,"0"),fprintf(report,"答案都错了。。。期望%d但输出%d",b,a);
}

没方案或方案错了就会得到3分233

贴一下没方案的代码

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "course"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0;rg char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x;
}
const int maxn=520,maxm=maxn;
int m,n,w[maxn],fir[maxn],dis[maxm],nxt[maxm],id;
il vd add(int a,int b){nxt[++id]=fir[a],fir[a]=id,dis[id]=b;}
int f[maxn][maxn];
il vd dp(int x,int M){
    if(M==0)return;
    erep(i,x){
	rep(j,1,M-1)f[dis[i]][j]=f[x][j];
	dp(dis[i],M-1);
	drep(j,M,1)f[x][j]=max(f[x][j],f[dis[i]][j-1]+w[dis[i]]);
    }
}
int main(){
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
    m=gi(),n=gi();
    rep(i,1,m)add(gi(),i),w[i]=gi();
    dp(0,n);
    printf("%d\n",f[0][n]);
    return 0;
}

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蛤蛤的烦恼

利用了凸多边形的性质。

然而窝不会证。

f[l][r]表示从l开始遍历l~r的最小代价。

g[l][r]表示从r开始遍历l~r的最小代价。

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#include<cmath>
#define Fname "frog"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0;rg char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x;
}
#define db double
#define maxn 721
db x[maxn],y[maxn];
db f[maxn][maxn],g[maxn][maxn];
il db dist(int a,int b){return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));}
int main(){
    #ifdef xzz
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
    #endif
    int n=gi(),r;
    rep(i,1,n)scanf("%lf%lf",&x[i],&y[i]);
    rep(siz,2,n)drep(l,n-siz+1,1){
	r=l+siz-1;
	f[l][r]=min(f[l+1][r]+dist(l,l+1),g[l+1][r]+dist(l,r));
	g[l][r]=min(g[l][r-1]+dist(r-1,r),f[l][r-1]+dist(l,r));
    }
    printf("%.3f\n",f[1][n]);
    return 0;
}

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警卫安排

灰常经典乐，以前早写过类似的辣

f[i]表示父亲染i，g表示i染i，h表示i的儿子染i

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "security"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0;rg char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x;
}
const int maxn=722;
int n,w[maxn];
int fir[maxn],dis[maxn],nxt[maxn],id;
il vd add(int a,int b){nxt[++id]=fir[a],fir[a]=id,dis[id]=b;}
int f[maxn],g[maxn],h[maxn];
//f父亲or自己染自己，g自己染自己，h儿子or自己染自己
il vd dp(int x){
    f[x]=0,g[x]=w[x],h[x]=1e9;
    int Sum=0;
    erep(i,x){
	dp(dis[i]);
	Sum+=h[dis[i]];
	g[x]+=f[dis[i]];
	f[x]+=h[dis[i]];
    }
    erep(i,x)h[x]=min(h[x],Sum-h[dis[i]]+g[dis[i]]);
    h[x]=min(h[x],g[x]);
    f[x]=min(f[x],g[x]);
}
int isson[maxn];
int main(){
    #ifdef xzz
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
    #endif
    n=gi();
    int son,s,a,root;
    rep(i,1,n){
	s=gi();
	w[s]=gi();
	son=gi();
	while(son--)a=gi(),isson[a]=1,add(s,a);
    }
    rep(i,1,n)if(!isson[i])root=i;
    dp(root);
    printf("%d\n",min(g[root],h[root]));
    return 0;
}

---

最长上升子序列

SBT。k前去掉比a[k]打的，k后去掉比a[k]小的，显然答案不变而且能取到k

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "lis"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0;rg char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x;
}
const int maxn=2e5+2;
int d[maxn],a[maxn];
int main(){
#ifdef xzz
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
#endif
    int s,n=gi(),k=gi(),maxf=0;
    rep(i,1,n)a[i]=gi();
    d[0]=-2e9;
    rep(i,1,n)d[i]=2e9;
    rep(i,1,n){
	if((i<k&&a[i]>=a[k])||(i>k&&a[i]<=a[k]))continue;
	s=lower_bound(d,d+maxf+1,a[i])-d;
	//f[i]=s
	d[s]=min(d[s],a[i]);maxf=max(maxf,s);
    }
    printf("%d\n",maxf);
    return 0;
}

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最短回文串

同编辑距离。稍作改动就行

f[l][r]表示[l,r]变成回文串最少插几个字符

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "palindrome"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0;rg char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x;
}
char s[5010];
short f[5010][5010];
int main(){
#ifdef xzz
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
#endif
    int n=gi(),r;
    scanf("%s",s+1);
    rep(siz,2,n)drep(l,n-siz+1,1){
	r=l+siz-1;
	if(s[l]==s[r])f[l][r]=f[l+1][r-1];
	else f[l][r]=min(f[l+1][r],f[l][r-1])+1;
    }printf("%d\n",f[1][n]);
    return 0;
}

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水表搞定了！

备份，密码同博客密码