poj 1151 Atlantis 求矩形面积并

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 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 

//

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn=110;
//坐标是浮点数
struct LINE
{
    double  x, y_down, y_up;
    int  flag;
    bool operator<(const LINE &a)const
    {
        return  x<a.x;
    }
}line[2*maxn];
struct TREE
{
    double  y_down, y_up;
    double  x;
    int          cover; //用以表示加进线段树中的线段次数
    bool      flag; //此标记用来表示是否有超元线段；为了处理方便加上去的
}tree[1000*maxn];
int    n;
double  x1, y1, x2, y2;
int          index=0;
double          y[2*maxn];
void build(int i, int l, int r)
{
       tree[i].x = -1; //-1表示该区间已经没有线段
       tree[i].cover = 0; //表示该区间上有多少条线段；左边线段加进去则++，右边线段加进去则--
       tree[i].y_down = y[l];
       tree[i].y_up = y[r];
       tree[i].flag = false;
       if(l+1==r)
       {
           tree[i].flag = true;
           return;
       }
       int mid=(l+r)>>1;
       build(2*i, l, mid);
       build(2*i+1, mid, r);
}
double insert(int i, double x, double l, double r, int flag) //flag表示为左边还是右边
{
    if (r<=tree[i].y_down || l>=tree[i].y_up) return 0;
    if ( l<=tree[i].y_down && tree[i].y_up <= r && tree[i].flag)
    {
        if (tree[i].cover > 0)
        {
             double temp_x = tree[i].x;
             double ans=(x-temp_x)*(tree[i].y_up - tree[i].y_down);
             tree[i].x = x;
             tree[i].cover += flag;
             return ans;
        }
        else
        {
            tree[i].cover += flag;
            tree[i].x = x;
            return 0;
        }
    }
    double ans1, ans2;
    ans1 = insert(2*i, x, l, r, flag);
    ans2 = insert(2*i+1, x, l, r, flag);
    return ans1+ans2;
}
int main( )
{
    int  count=0;
    while (scanf("%d", &n)!=EOF&&n)
    {
        index = 1;
        for (int i=1; i<=n; i++)
        {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            y[index] = y1;
            line[index].x = x1;
            line[index].y_down = y1;
            line[index].y_up = y2;
            line[index].flag = 1; //1表示左边
            index++;
            y[index] = y2;
            line[index].x = x2;
            line[index].y_down = y1;
            line[index].y_up = y2;
            line[index].flag = -1; //-1表示右边
            index++;
        }
        sort(&y[1], &y[index]); //把所有的纵坐标按从小到大排序,把1写成了0，WA一次
        sort(&line[1], &line[index]);
        build(1, 1, index-1);
        double ans=0;
        for (int i=1;i<index; i++)
        {
            ans += insert(1, line[i].x, line[i].y_down, line[i].y_up, line[i].flag);
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n", ++count, ans);
    }
    return 0;
}