前缀和（二维）

T1P2280 [HNOI2003] 激光炸弹

二维前缀和模板题，坑点：坐标+1防止数组越界。范围变成（1，N+1）

#include <bits/stdc++.h>
using namespace std;
const int N = 5001;
int a[5010][5010];
int n, m, ans;
int main() {
	memset(a, 0, sizeof(a));
	cin >> n >> m;
	for (int i = 1; i <= n; i ++) {
		int x, y, v;
		cin >> x >> y >> v;
		a[x + 1][y + 1] += v; 
	}
	for (int i = 1; i <= N; i ++)
		for (int j = 1; j <= N; j ++)
			a[i][j] += a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
	for (int i = m; i <= N; i ++) {
		for (int j = m; j <= N; j ++) {
			int val;
			val = a[i][j] - a[i - m][j] - a[i][j - m] + a[i - m][j - m];
			ans = max(ans, val);
		}
	}
	cout << ans;
}

T2P1719 最大加权矩形

注意到n=120，数据很小，可以直接考虑稍微暴力的解法，枚举子矩阵的行数和列数，再枚举子矩阵右下角的坐标，复杂度n的4次方。

#include <bits/stdc++.h>
using namespace std;
int n, a[130][130];
int main() {
	cin >> n;
	for (int i = 1; i <= n; i ++) {
		for (int j = 1; j <= n; j ++) {
			cin >> a[i][j];
			a[i][j] += a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
		}
	}
	int ans = -0x3f3f3f3f;
	for (int r = 1; r <= n; r ++) {//枚举子矩阵的行数 
		for (int c = 1; c <= n; c ++) {//列数 
			for (int i = r; i <= n; i ++) {
				for (int j = c; j <= n; j ++) {
					int val;
					val = a[i][j] - a[i - r][j] - a[i][j - c] + a[i - r][j - c];
					ans = max(ans, val);
				}
			} 
		}
	}
	cout << ans;
}