2 二维(矩阵)差分
核心:将a[] 一个矩阵区域的+c 映射为b[]四个边界点的+或-c,最后对b[]求一次前缀和即可得到a进行若干次加矩阵后的输出;
#include <iostream>
using namespace std;
const int N = 1010;
int a[N][N], b[N][N];
int n, m, q;
void insert(int x1, int y1, int x2, int y2, int c){
b[x1][y1] += c;
b[x2 + 1][y1] -= c;
b[x1][y2 + 1] -= c;
b[x2 + 1][y2 + 1] += c;
}
int main(){
scanf("%d%d%d", &n, &m, &q);
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
scanf("%d", &a[i][j]);
insert(i, j, i, j, a[i][j]);
}
}
while(q--){
int x1, y1, x2, y2, c;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &c);
insert(x1, y1, x2, y2, c);
}
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("%d ", b[i][j]);
}
puts("");
}
}