Problem Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the
total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

 

Input

The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000),
not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.



The input file is terminated by a line containing a single 0. Don’t process it.

 

Output

For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area
(i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.



Output a blank line after each test case.

 

Sample Input

2

10 10 20 20

15 15 25 25.5

0

 

Sample Output

Test case #1

Total explored area: 180.00 
所谓的离散化，大家能够简单的理解为，将一组非常大的数据，浓缩为一组非常小的数据，用这组数据来取代原数据的作用，
比方给你1000个数，数的范围为(1,1e18)我们这里就能够用离散化，因为仅仅有1000个数。我们能够用一个数组的下标代表提供的每一数。假设须要这个数据了，因为是下标，能够直接通过下标获得，如此就是离散化。
扫描线的知识提供一个大牛的博客：http://www.cnblogs.com/scau20110726/archive/2013/04/12/3016765.html

讲得非常基础，是个非常不错的博客
然后提醒一下这个扫描线要注意的问题就是区间的问题
一般的线段树以及我们的区间改动合并。都有一个共同点，就是不会出现区间缺失的现象，什么叫区间缺失。顾名思义，区间缺失就是缺少一些区间没有进行运算，这里的扫描线就会遇到这个问题。
普遍的，我们的线段树以及数据区间分布是这种：
[1, a][a + 1, b][b + 1, c][c + 1, d][d + 1, e].......
可是假设仅仅是简简单单的用这个来解决扫描线的问题会导致错误，为什么因为。他没有涉及到[a,a + 1]，在扫描线中会出现[a,a + 1]中的数据，而经常使用的线段树的区间概念是无法解决这种问题的，出现了所谓的区间缺失，如何解决，以下的代码给出了解决方式，这里简单的提一下，就是利用[ , ),这个区间性质，左闭右开。就可以解决区间缺失问题


#include <cstdio>

#include <cstring>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

typedef long long LL;

#define lson rt << 1, l, mid

#define rson rt << 1|1, mid + 1, r

const int MAXN = 2000 + 5;

int  Col[MAXN << 2], n, cnt, res;

double X[MAXN << 2], Sum[MAXN << 2];

struct seg {

    double l,r,h;

    int s;

    seg() {}

    seg(double l,double r,double h,int s):l(l),r(r),h(h),s(s) {}

    bool operator < (const seg & object) const {

        return h < object.h;

    }

} S[MAXN];

void pushup(int rt,int l,int r) {

    if (Col[rt]) Sum[rt] = X[r+1] - X[l];//利用[ , ),这个区间性质。左闭右开

    else if (l == r) Sum[rt] = 0;

    else Sum[rt] = Sum[rt<<1] + Sum[rt<<1|1];

}

void update(int L, int R, int c,int rt,int l, int r) {

    if(L <= l && r <= R) {

        Col[rt] += c;

        pushup(rt,l,r);

        return ;

    }

    int mid = (l + r) >> 1;

    if(L <= mid) update(L, R, c, lson);

    if(R > mid) update(L, R, c, rson);

    pushup(rt,l,r);

}

int binary_find(double x){

    int lb = -1,ub = res - 1;

    while(ub - lb > 1){

        int mid = (lb + ub) >> 1;

        if(X[mid] >= x) ub = mid;

        else lb = mid;

    }

    return ub;

}

int main() {

    int cas = 1;

    while(~ scanf("%d", &n), n) {

        cnt = res = 0;

        for(int i = 0 ; i < n; i ++) {

            double a,b,c,d;

            scanf("%lf%lf%lf%lf",&a, &b, &c,&d);

            S[cnt] = seg(a, c, b, 1);

            X[cnt ++] = a;

            S[cnt] = seg(a, c, d, -1);

            X[cnt ++] = c;

        }

        sort(X, X + cnt);

        sort(S, S + cnt);

        res ++;

        for(int i = 1; i < cnt; i ++) {

            if(X[i] != X[i - 1]) X[res ++] = X[i];

        }

        memset(Sum, 0, sizeof(Sum));

        memset(Col, 0, sizeof(Col));

        double ans = 0;

        for(int i = 0;i < cnt - 1;i ++){

            int l = binary_find(S[i].l);

            int r = binary_find(S[i].r) - 1;//利用[ , ),这个区间性质，左闭右开

            update(l, r, S[i].s, 1, 0, res - 1);

            ans += Sum[1] * (S[i + 1].h - S[i].h);

        }

        printf("Test case #%d\nTotal explored area: %.2lf\n\n",cas++ , ans);

    }

    return 0;

}