可能有多组数据，读到n=0为止(不超过15组)

每组数据第一行一个数n，表示矩形个数(n<=100)

接下来n行每行4个实数x1,y1,x2,y1(0 <= x1 < x2 <= 100000;0 <= y1 < y2 <= 100000)，表示矩形的左下角坐标和右上角坐标

每组数据输出一行表示答案

2
10 10 20 20
15 15 25 25.5
0

180.00

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;int read() {int x = 0, f = 1; char c = getchar();while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }return x * f;
}#define maxn 110struct Line {int l, r, h, tp;Line() {}Line(int _1, int _2, int _3, int _4): l(_1), r(_2), h(_3), tp(_4) {}bool operator < (const Line& t) const { return h < t.h; }
} ls[maxn<<1];
double posx[maxn<<1], posy[maxn<<1], numx[maxn<<1], numy[maxn<<1], ans;int cntv[maxn<<3];
double sumv[maxn<<3];
void maintain(int L, int R, int o) {int lc = o << 1, rc = lc | 1;if(cntv[o]) sumv[o] = numx[R] - numx[L-1];else if(L == R) sumv[o] = 0;else sumv[o] = sumv[lc] + sumv[rc];return ;
}
void update(int L, int R, int o, int ql, int qr, int v) {if(ql <= L && R <= qr) {cntv[o] += v;return maintain(L, R, o);}int M = L + R >> 1, lc = o << 1, rc = lc | 1;if(ql <= M) update(L, M, lc, ql, qr, v);if(qr > M) update(M+1, R, rc, ql, qr, v);return maintain(L, R, o);
}int main() {while(1) {int n = read(), cntx = 0, cnty = 0, cntl = 0;if(!n) break;for(int i = 1; i <= n; i++) {double x1, x2, y1, y2;scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);posx[++cntx] = x1; posx[++cntx] = x2;posy[++cnty] = y1; posy[++cnty] = y2;numx[cntx-1] = posx[cntx-1]; numx[cntx] = posx[cntx];numy[cnty-1] = posy[cnty-1]; numy[cnty] = posy[cnty];}sort(numx + 1, numx + cntx + 1);sort(numy + 1, numy + cnty + 1);for(int i = 1; i <= n; i++) {int x1, x2, y1, y2;x1 = lower_bound(numx + 1, numx + cntx + 1, posx[(i<<1)-1]) - numx;x2 = lower_bound(numx + 1, numx + cntx + 1, posx[i<<1]) - numx;y1 = lower_bound(numy + 1, numy + cnty + 1, posy[(i<<1)-1]) - numy;y2 = lower_bound(numy + 1, numy + cnty + 1, posy[i<<1]) - numy;ls[++cntl] = Line(x1, x2, y1, 1);ls[++cntl] = Line(x1, x2, y2, -1);}sort(ls + 1, ls + cntl + 1);memset(cntv, 0, sizeof(cntv));memset(sumv, 0, sizeof(sumv));ans = 0;double start = numx[1];for(int i = 1; i < cntx; i++) numx[i] = numx[i+1] - start;for(int i = 1; i < cntl; i++) {if(ls[i].l < ls[i].r) update(1, cntx - 1, 1, ls[i].l, ls[i].r - 1, ls[i].tp);ans += (numy[ls[i+1].h] - numy[ls[i].h]) * sumv[1];}printf("%.2lf\n", ans);}return 0;
}