LG4158 "SCOI2009" painters linear DP

Problem Description

LG4158

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answer

Set \ (opt [i] [j ] [k] \) representative of the \ ((i, k) \ ) brush \ (J \) the number of times the program.

DP start sequence a little problem, tune for a long time.

Be sure to think carefully DP order problem

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\(\mathrm{Code}\)

#include<bits/stdc++.h>
using namespace std;

template <typename Tp>
void read(Tp &x){
    x=0;char ch=1;int fh;
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-'){
        fh=-1;ch=getchar();
    }
    else fh=1;
    while(ch>='0'&&ch<='9'){
        x=(x<<1)+(x<<3)+ch-'0';
        ch=getchar();
    }
    x*=fh;
}

const int maxn=53;

int n,m,t;
int a[maxn][maxn],s[maxn][maxn*maxn],opt[maxn][maxn*maxn][maxn];
int f[maxn][maxn*maxn];


void println_opt(){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            for(int k=0;k<=t;k++){
                printf("opt[%d][%d][%d]=%d\n",i,j,k,opt[i][j][k]);
            }
        }
    }
}

int main(){
    read(n);read(m);read(t);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%1d",&a[i][j]);
            s[i][j]=s[i][j-1]+a[i][j];
        }
    }
//  opt[1][0][0]=1;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
//          int ed=min((i-1)*n+j,t);
            for(int k=1;k<=m;k++){
//              opt[i][j][k]=opt[i][0][k-1]+max(s[i][k],k-s[i][k]);
                for(int p=j-1;p<k;p++){
                    opt[i][j][k]=max(opt[i][j][k],opt[i][j-1][p]+max(s[i][k]-s[i][p],k-p-(s[i][k]-s[i][p])));
                }
            }
        }
        //for(int j=1;j<=t;j++) opt[i+1][0][j]=opt[i][m][j];
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=t;j++){
            for(int k=0;k<=min(j,m);k++){
                f[i][j]=max(f[i][j],f[i-1][j-k]+opt[i][k][m]);
            }
        }
    }
    int ans=0;
    for(int i=1;i<=t;i++) ans=max(ans,f[n][i]);
    printf("%d\n",ans);
//  println_opt();
    return 0;
}