hdu1542 Atlantis （线段树|扫描线）

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11253    Accepted Submission(s): 4778

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

题目大意：给出n个矩形（左上角和右下角两个坐标），求总的矩形面积，重复的部分只算一次。
思路：线段树+离散化+扫描线。
#include<cstdio> 
#include<cstring> 
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1000005;
int mark[maxn<<2];
double sum[maxn<<2],thash[maxn];
struct node
{
	double l,r,h;
	int cover;
	node(){}
	node(double a,double b,double c,int d):l(a),r(b),h(c),cover(d){}
	bool operator <(const node&x)const
	{
		return h<x.h;
	}
}s[maxn<<2];
void updatefather(int pos,int l,int r)
{
	if(mark[pos])
	{		
		sum[pos]=thash[r+1]-thash[l];
	}
	else if(l==r)sum[pos]=0;//变成一个点，所以没长度
	else sum[pos]=sum[pos<<1]+sum[pos<<1|1];
}
void update(int pos,int l,int r,int ql,int qr,int value)
{
	if(ql<=l&&r<=qr)
	{	
		mark[pos]+=value;
		updatefather(pos,l,r);
		return;
	}
	int mid=(l+r)>>1;
	if(ql<=mid)update(pos<<1,l,mid,ql,qr,value);
	if(qr>mid)update(pos<<1|1,mid+1,r,ql,qr,value);
	updatefather(pos,l,r);
}
int main()
{
	int n,cas=1;
	while(scanf("%d",&n)==1&&n)
	{
		memset(mark,0,sizeof(mark));
		int m=0,num=2;
		double x1,x2,y1,y2,ans=0;
		for(int i=0;i<n;i++)
		{
			scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
			s[m]=node(x1,x2,y1,1);
			thash[++m]=x1;
			s[m]=node(x1,x2,y2,-1);
			thash[++m]=x2;
		}
		printf("Test case #%d\n",cas++);
		sort(s,s+m);
		sort(thash+1,thash+1+m);
		for(int i=2;i<=m;i++)if(thash[i]!=thash[i-1])thash[num++]=thash[i];
		for(int i=0;i<m;i++)
		{
			int L=lower_bound(thash+1,thash+num,s[i].l)-thash;
			int R=lower_bound(thash+1,thash+num,s[i].r)-thash-1; 
			if(i)ans+=sum[1]*(s[i].h-s[i-1].h);
			update(1,1,num-1,L,R,s[i].cover);
		}
		printf("Total explored area: %.2lf\n\n",ans);
	}
}