luoguP4158 [SCOI2009]粉刷匠

Description

連結

Solution

先一個DP處理一波每一行

然後再做一個揹包

#include<bits/stdc++.h>

using namespace std;

inline int read()
{
    int f = 1,x = 0;
    char ch;
    do
    {
        ch = getchar();
        if(ch == '-') f = -1;
    }while(ch < '0'||ch > '9');
    do
    {
        x 
 
= (x<<3) + (x<<1) + ch - '0';
        ch = getchar();
    }while(ch >= '0'&&ch <= '9');
    return f*x;
}

int n,m,t;
char a[50 + 10];
int sum[50 + 10][50 + 10];
int dp[50 + 10][50 + 10][2500 + 10];
int ans[50+ 10][2500 + 10];

int main()
{
    n = read(),m = read(),t = read();
    for(int i=1
 
;i<=n;i++)
    {
        scanf("%s",a+1);
        for(int j=1;j<=m;j++)
        {
            sum[i][j] = sum[i][j-1] + a[j] - '0';
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            for(int k=1;k<=min(t,j);k++)
            {
                
 
for(int h=j-1;h>=k-1;h--)
                {
                    dp[i][j][k] = max(dp[i][j][k] ,dp[i][h][k-1] + max(sum[i][j] - sum[i][h],j - h-(sum[i][j] - sum[i][h])));    
                } 
            }
                
        }
    }
    int ans_M = 0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=t;j++)
        {
            for(int k=1;k<=min(j,m);k++)
                ans[i][j] = max(ans[i][j],ans[i-1][j-k] + dp[i][m][k]);
            if(i==n) ans_M = max(ans_M , ans[i][j]);
        }
    }
    cout << ans_M << endl;
}