ACM POJ1151 (HDU 1542) Atlantis（矩形面积并，线段树+离散化+扫描线）

Atlantis

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9693		Accepted: 3791

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 

Source

Mid-Central European Regional Contest 2000

 

 

/*
POJ 1151 Atlantis
求矩形并的面积（线段树+离散化） 
*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 201
struct Node
{
    int l,r;//线段树的左右整点
    int c;//c用来记录重叠情况
    double cnt,lf,rf;//
    //cnt用来计算实在的长度，rf,lf分别是对应的左右真实的浮点数端点 
}segTree[MAXN*3];     
struct Line
{
    double x,y1,y2;
    int f;
}line[MAXN];
//把一段段平行于y轴的线段表示成数组 ，
//x是线段的x坐标，y1,y2线段对应的下端点和上端点的坐标 
//一个矩形 ，左边的那条边f为1，右边的为-1，
//用来记录重叠情况，可以根据这个来计算，nod节点中的c 

bool cmp(Line a,Line b)//sort排序的函数
{
    return a.x < b.x;
}     

double y[MAXN];//记录y坐标的数组
void Build(int t,int l,int r)//构造线段树
{
    segTree[t].l=l;segTree[t].r=r;
    segTree[t].cnt=segTree[t].c=0;
    segTree[t].lf=y[l];
    segTree[t].rf=y[r];
    if(l+1==r)  return;
    int mid=(l+r)>>1;
    Build(t<<1,l,mid);
    Build(t<<1|1,mid,r);//递归构造 
}     
void calen(int t)//计算长度
{
    if(segTree[t].c>0)
    {
        segTree[t].cnt=segTree[t].rf-segTree[t].lf;
        return;
    }    
    if(segTree[t].l+1==segTree[t].r)  segTree[t].cnt=0;
    else  segTree[t].cnt=segTree[t<<1].cnt+segTree[t<<1|1].cnt;
}     
void update(int t,Line e)//加入线段e，后更新线段树
{
    if(e.y1==segTree[t].lf&&e.y2==segTree[t].rf)
    {
        segTree[t].c+=e.f;
        calen(t);
        return;
    }    
    if(e.y2<=segTree[t<<1].rf)  update(t<<1,e);
    else if(e.y1>=segTree[t<<1|1].lf)  update(t<<1|1,e);
    else
    {
        Line tmp=e;
        tmp.y2=segTree[t<<1].rf;
        update(t<<1,tmp);
        tmp=e;
        tmp.y1=segTree[t<<1|1].lf;
        update(t<<1|1,tmp);
    }    
    calen(t);
}    
int main()
{
    int i,n,t,iCase=0;
    double x1,y1,x2,y2;
    while(scanf("%d",&n),n)
    {
        iCase++;
        t=1;
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[t].x=x1;
            line[t].y1=y1;
            line[t].y2=y2;
            line[t].f=1;
            y[t]=y1;
            t++;
            line[t].x=x2;
            line[t].y1=y1;
            line[t].y2=y2;
            line[t].f=-1;
            y[t]=y2;
            t++;
        } 
        sort(line+1,line+t,cmp);
        sort(y+1,y+t);   
        Build(1,1,t-1);
        update(1,line[1]);
        double res=0;
        for(i=2;i<t;i++)
        {
            res+=segTree[1].cnt*(line[i].x-line[i-1].x);
            update(1,line[i]);
        }    
        printf("Test case #%d\nTotal explored area: %.2f\n\n",iCase,res);
        //看来POJ上%.2f可以过，%.2lf却不行了 
    }    
    return 0;
}