HDU 1542 Atlantis

题目意思：就是说一个平面上有很多个矩形，求他们在平面上的覆盖面积

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 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>
#include<cmath>
using namespace std;

#define maxn 1005

struct node
{
    int L, R, cover;
    int Mid(){return (L+R)/2;}
};
struct point
{
    double x, y1, y2;
    bool IsLeft;
};
bool cmp(point a, point b)
{
    return a.x < b.x;
}

node a[maxn];
double Len, f[maxn];

void BuildTree(int r, int L, int R);
void Insert(int r, int L, int R, bool IsLeft);
void Query(int r);

int main()
{
    int N, ncase = 1;

    while(scanf("%d", &N), N)
    {
        int i, np=0, nf=0;
        double S = 0, x1, y1, x2, y2;
        point p[maxn];

        for(i=0; i<N; i++)
        {
            cin >> x1 >> y1 >> x2 >> y2;
            p[np].x = x1, p[np].y1 = y1, p[np].y2 = y2, p[np++].IsLeft = true;
            p[np].x = x2, p[np].y1 = y1, p[np].y2 = y2, p[np++].IsLeft = false;

            f[nf++] = y1, f[nf++] = y2;
        }

        sort(f, f+nf);
        sort(p, p+np, cmp);

        nf = unique(f, f+nf) - f;

        BuildTree(1, 0, nf-1);

        for(i=0; i<np-1; i++)
        {
            int L = lower_bound(f, f+nf, p[i].y1) - f;
            int R = lower_bound(f, f+nf, p[i].y2) - f;

            Insert(1, L, R, p[i].IsLeft);

            Len = 0;
            Query(1);

            S += (p[i+1].x-p[i].x) * Len;
        }

        printf("Test case #%d\n", ncase++);
        printf("Total explored area: %.2f\n\n", S);
    }

    return 0;
}

void BuildTree(int r, int L, int R)
{
    a[r].L = L, a[r].R = R, a[r].cover = false;

    if(L + 1 == R)return ;

    BuildTree(r*2, L, a[r].Mid());
    BuildTree(r*2+1, a[r].Mid(), R);
}
void Insert(int r, int L, int R, bool IsLeft)
{

    if(a[r].L == L && a[r].R == R)
    {
        if(IsLeft)
            a[r].cover++;
        else
            a[r].cover--;

        return ;
    }

    if(R <= a[r].Mid())
        Insert(r*2, L, R, IsLeft);
    else if(L >= a[r].Mid())
        Insert(r*2+1, L, R, IsLeft);
    else
    {
        Insert(r*2, L, a[r].Mid(), IsLeft);
        Insert(r*2+1, a[r].Mid(), R, IsLeft);
    }
}
void Query(int r)
{
    if(a[r].cover)
    {
        Len += f[a[r].R] - f[a[r].L];
        return ;
    }

    if(a[r].L+1 == a[r].R)return ;

    Query(r*2);
    Query(r*2+1);
}