给出矩形的 对角线上的两个点的坐标 (x11, y11)， (x12, y12)， (x21, y21), (x22, y22)。

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int x01 = max(min(x11, x12), min(x21, x22));
int x02 = min(max(x11, x12), max(x21, x22));
int y01 = max(min(y11, y12), min(y21, y22));
int y02 = min(max(y11, y12), max(y21, y22));
if (x01 < x02 && y01 < y02){
    // 说明这两个矩形相交
    int s = (x02 - x01) * (y02 - y01); // 相交的面积
}
else{ // 这两个不相交
    
}

例题 ：

题目页面1
题目页面2

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#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 5;
const int mod = 998244353;
struct node{
    int x1, y1, x2, y2;
    ll sum, s;
}arc[5005];
int x01, x02, y01, y02;
int main(){
    int t, n;
    cin >> t;
    while(t--){
        cin >> n;
        for (int i = 1; i <= n; i++){
            arc[i].sum= 0;
            arc[i].s = 0;
            cin >> arc[i].x1 >> arc[i].y1 >> arc[i].x2 >> arc[i].y2;
            for (int j = 1; j < i; j++){
                x01 = max(min(arc[i].x1, arc[i].x2), min(arc[j].x1, arc[j].x2));
                x02 = min(max(arc[i].x1, arc[i].x2), max(arc[j].x1, arc[j].x2));
                y01 = max(min(arc[i].y1, arc[i].y2), min(arc[j].y1, arc[j].y2));
                y02 = min(max(arc[i].y1, arc[i].y2), max(arc[j].y1, arc[j].y2));
                if (x01 < x02 && y01 < y02){
                    arc[i].sum++;
                    arc[i].s += (x02 - x01) * (y02 - y01);
                    arc[j].sum++;
                    arc[j].s += (x02 - x01) * (y02 - y01);
                }
            }
        }
        for (int i = 1; i <= n; i++){
            cout << arc[i].sum << ' ' << arc[i].s << "\n";
        }

    }
    return 0;
}