【题解】LuoGu4158：粉刷匠

原题传送门
很好的一个dp
令 $f_{i,j}$表示前 $i$行刷了 $j$次的答案， $g_{i,j,k}$表示第 $i$行前 $k$个，刷了 $j$次的最大价值

• $f_{i,j}=max(f_{i-1,j-k}+g_{i,k,m})$
• $g_{i,j,k}=max(g_{i,j-1,l}+max(sum_{i,k}-sum_{i,l},(k-l)-(sum_{i,k}-sum_{i,l})))$

COde：

#include <bits/stdc++.h>
#define maxn 60
#define maxm 2510
using namespace std;
int n, m, t, f[maxn][maxm], g[maxn][maxn][maxn], sum[maxn][maxn];
char s[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

int main(){
	scanf("%d%d%d", &n, &m, &t);
	for (int i = 1; i <= n; ++i){
		scanf("%s", s + 1);
		for (int j = 1; j <= m; ++j) sum[i][j] = sum[i][j - 1] + (s[j] == '1') ;
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= min(m, t); ++j)
			for (int k = 1; k <= m; ++k)
				for (int l = j - 1; l < k; ++l)
					g[i][j][k] = max(g[i][j][k], g[i][j - 1][l] + max(sum[i][k] - sum[i][l], k - l - sum[i][k] + sum[i][l]));
	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] + g[i][k][m]);
	int ans = 0;
	for (int i = 0; i <= t; ++i) ans = max(ans, f[n][i]);
	printf("%d\n", ans);
	return 0;
}