AcWing 798. 差分矩阵
原题链接
简单
作者:
我要出去乱说
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2021-02-06 20:34:12
,
所有人可见
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阅读 363
#include <iostream>
using namespace std;
const int N = 1010;
int a[N][N], b[N][N]; //a储存原矩阵,b储存差分矩阵
void insert(int x1, int y1, int x2, int y2, int c) //差分代码
{
b[x1][y1] += c;
b[x2 + 1][y1] -= c;
b[x1][y2 + 1] -= c;
b[x2 + 1][y2 + 1] += c;
}
int main()
{
int n, m, q;
cin >> n >> m >> q;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
{
cin >> a[i][j];
insert(i, j, i, j, a[i][j]); //输入a同时构造差分矩阵b
}
while (q--)
{
int x1, y1, x2, y2, c;
cin >> 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]; //从差分矩阵b还原到前缀和矩阵a
printf("%d ", b[i][j]);
}
puts("");
}
return 0;
}