题解 | #【模板】二维差分#第2版

#include <iostream>
typedef long long ll;
using namespace std;
ll n, m, q,  b[1005][1005], x1, y1, x2, y2, k, tmp;
void diff(int x1, int y1, int x2, int y2, int k) {
    b[x1][y1] += k;
    b[x1][y2 + 1] -= k;
    b[x2 + 1][y1] -= k;
    b[x2 + 1][y2 + 1] += k;
}
int main() {
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            scanf("%lld", &tmp);//将矩阵原来的元素也视为一次差分操作，那么矩阵原来的元素都为0
            diff(i, j, i, j, tmp);
        }
    }
    while (q--) {
        scanf("%lld %lld %lld %lld %lld", &x1, &y1, &x2, &y2, &k);
        diff(x1, y1, x2, y2, k);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];
            printf("%lld ", b[i][j]);
        }
        printf("\n");
    }
}