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28 lines
885 B
28 lines
885 B
#include <bits/stdc++.h>
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using namespace std;
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const int N = 1010;
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int b[N][N], s[N][N];
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int n, m;
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int main() {
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cin >> n >> m;
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while (m--) {
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// 从0开始构建差分数组
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int x1, y1, x2, y2;
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cin >> x1 >> y1 >> x2 >> y2;
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b[x1][y1] += 1; // 进行子矩阵的加减，差分
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b[x2 + 1][y1] -= 1;
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b[x1][y2 + 1] -= 1;
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b[x2 + 1][y2 + 1] += 1;
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}
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// 还原为原始数组
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for (int i = 1; i <= n; i++) {
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for (int j = 1; j <= n; j++) {
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s[i][j] = b[i][j] + s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1]; // 把之前的加减结果进行求和
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printf("%d ", s[i][j]); // 注意输出格式，每个数带一个空格
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}
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printf("\n"); // 结束一行的输出输出一个换行符号
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}
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return 0;
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} |