求n个矩阵面积相交的部分，和求面积并一样，不过这里需要开两个数组保存覆盖一次和覆盖两次以上的次数的部分，还是模板，主要注意点就是pushup部分，如果我已经被两次覆盖，那我的两个数组在这个root点的宽度就可以直接算了，如果我被一次覆盖，那么我一个覆盖的部分可以直接计算，两次覆盖的部分取决于sum1数组，如果我不是一个线段，那就是0，如果我是一个线段，那么我就是由左边区间覆盖一次和右边区间覆盖一次相加得来。如果我这个区间现在没有被覆盖到，那么我的两个数组都是由各自部分的区间和得来的
#include<map> #include<set> #include<ctime> #include<cmath> #include<stack> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define first fi #define second se #define lowbit(x) (x & (-x))typedef unsigned long long int ull; typedef long long int ll; const double pi = 4.0*atan(1.0); const int inf = 0x3f3f3f3f; const int maxn = 2005; const int maxm = 305; using namespace std;int n, m, tol, T; struct Node{double l, r, h;int f;bool operator < (Node a) const {return h < a.h;} }; Node node[maxn]; double a[maxn]; int cnt[maxn << 2]; double sum1[maxn << 2]; double sum2[maxn << 2];void init() {memset(a, 0, sizeof a);memset(cnt, 0, sizeof cnt);memset(sum1, 0, sizeof sum1);memset(sum2, 0, sizeof sum2);memset(node, 0, sizeof node); }void pushup(int left, int right, int root) {if(cnt[root] >= 2) {sum2[root] = sum1[root] = a[right+1] - a[left];} else if(cnt[root] == 1) {sum1[root] = a[right+1] - a[left];if(left == right) sum2[root] = 0;else sum2[root] = sum1[root << 1] + sum1[root << 1 | 1];} else {if(left == right) sum1[root] = sum2[root] = 0;else {sum1[root] = sum1[root << 1] + sum1[root << 1 | 1];sum2[root] = sum2[root << 1] + sum2[root << 1 | 1];}} }void update(int left, int right, int prel, int prer, int val, int root) {if(prel <= left && right <= prer) {cnt[root] += val;pushup(left, right, root);return ;}int mid = (left + right) >> 1;if(prel <= mid) update(left, mid, prel, prer, val, root << 1);if(prer > mid) update(mid+1, right, prel, prer, val, root << 1 | 1);pushup(left, right, root);return ; }int main() {scanf("%d", &T);while(T--) {init();scanf("%d", &n);double x1, y1, x2, y2;for(int i=1; i<=n; i++) {scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);node[2*i].l = node[2*i-1].l = x1;node[2*i].r = node[2*i-1].r = x2;node[2*i].h = y1, node[2*i-1].h = y2;node[2*i].f = 1, node[2*i-1].f = -1;a[2*i] = x1, a[2*i-1] = x2;}n <<= 1;sort(node+1, node+1+n);sort(a+1, a+1+n);int nn = unique(a+1, a+1+n) - (a+1);double ans = 0.0;for(int i=1; i<n; i++) {int l = lower_bound(a+1, a+1+nn, node[i].l) - a;int r = lower_bound(a+1, a+1+nn, node[i].r) - a;update(1, nn, l, r-1, node[i].f, 1);ans += sum2[1] * (node[i+1].h - node[i].h);}printf("%.2f\n", ans);}return 0; }