Codeforces Round #745 (Div. 2) C - Portal

题目链接：https://codeforces.com/contest/1581/problem/C

单纯的二维前缀和只能O(n²m²)，看了答案发现需要取后缀最小值，因为可以做到前后两部分互不干扰，时间复杂度为O(n²m)。

#include 
using namespace std;

const int N = 400 + 5; 

int t, n, m;

char g[N][N];
int sum[N][N], f[N];

int getSum(int x1, int y1, int x2, int y2) {
	return sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
}



int main(void) {
//	freopen("in.txt", "r", stdin);
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) {
			scanf("%s", g[i] + 1);
			for (int j = 1; j <= m; j++) g[i][j] -= '0';
		}
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++)
				sum[i][j] = sum[i][j - 1] + g[i][j];
		for (int j = 1; j <= m; j++)
			for (int i = 1; i <= n; i++)
				sum[i][j] += sum[i - 1][j];
		int ans = n * m;
		for (int i = 1; i <= n; i++)
			for (int j = i + 4; j <= n; j++) {
				for (int r = 4; r<= m; r++) {
					f[r] = getSum(i + 1, 1, j - 1, r - 1); // empty相关部分 
					f[r] -= getSum(i, 1, i, r - 1) + getSum(j, 1, j, r - 1); // 上下边相关部分 
					f[r] += 2 * (r - 1); // 上下边相关部分 
					f[r] += (j - i - 1) - getSum(i + 1, r, j - 1, r); // 右边贡献 
				}
				for (int r = m - 1; r >= 4; r--)
					f[r] = min(f[r], f[r + 1]);
				for (int l = 1; l + 3 <= m; l++) {
					int cur = f[l + 3];
					cur -= getSum(i + 1, 1, j - 1, l); // empty相关部分
					cur += getSum(i, 1, i, l) + getSum(j, 1, j, l); // 上下边相关部分 
					cur -= 2 * l; // 上下边相关部分 
					cur += (j - i - 1) - getSum(i + 1, l, j - 1, l); // 左边贡献 
					ans = min(ans, cur);
				}
			}
		printf("%dn", ans);
	}
	return 0;
}