题目描述

样例

源代码

#include <iostream>
#include <cstring>

using namespace std;

const int N = 100;

int flag[N + 1][N + 1];

int main()
{
    
    int n, x1, y1, x2, y2, sum = 0;

    // 变量初始化
    memset(flag, 0, sizeof(flag));

    // 输入数据,计算处理
    cin >> n;
    for (int i = 1; i <= n; i++) {
    
        // 输入数据
        cin >> x1 >> y1 >> x2 >> y2;

        // cumulative area
        sum += (x2 - x1) * (y2 - y1);

        // Mark and remove overlaps
        for (int i = x1; i < x2; i++)
            for (int j = y1; j < y2; j++) {
    
                if (flag[i][j])
                    sum--;
                flag[i][j] = 1;
            }
    }

    // 输出结果
    cout << sum << endl;

    return 0;
}

关于这题

Because a grid has four points when counting repetitions It's easy to overcount
So here we use the number of points instead of the number of grids

（Boxed with squares are the marked points Circles represent repetitions need to be reduced）