算法基础课题解合集

二维差分公式

b[x1][y1] += c;
b[x2 + 1][y1] -= c;
b[x1][y2 + 1] -= c;
b[x2 + 1][y2 + 1] += c;

这个公式是从二维前缀和反推过来的，因为差分是前缀和的逆运算。

代码

#include <iostream>
using namespace std;

const int N = 1010;
int n, m, q;
int a[N][N], b[N][N];

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()
{
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++)
            cin >> a[i][j];
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++)
            insert(i, j, i, j, a[i][j]);
    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];
    for (int i = 1; i <= n; i ++, cout << endl)
        for (int j = 1; j <= m; j ++)   
            cout << b[i][j] << " ";
    return 0;
}

这篇有点水啊……主要是懒了……就摸一下吧，hh

$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\mathcal{writer\enspace by \enspace acwing}$ : $\mathfrak{天元之弈}$