Atlantis HDU - 1542（扫描线）

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 

这是一个简单的线段树扫描线的应用

扫描线其实是个很神奇的东西，关于扫描线的具体描述，网上有很多的博客有讲解，在这里我就不献丑了，对于这个题，可以很快想到就是扫描线的应用，唯一需要注意的一些细节就是：

1、要对于数据进行离散化。

2、由于现在节点l,r对应的是一段区间，所以要先减一在加一。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4+10;

int n,mlen;
double msp[maxn];
double ans;

inline int find(double x){
	return lower_bound(msp,msp+mlen,x)-msp;
}
struct Line{
	double lx,rx,h;
	int cove;
	Line(){};
	Line(double _lx,double _rx,double _h,int _cove){
		lx = _lx,rx = _rx,h = _h,cove = _cove;
	}
	bool operator < (const Line &S)const{
		return h<S.h;
	}
}line[maxn<<1];

struct Info{
	int l,r,lazy;
	double sum;
};
struct SGT{
	Info tree[maxn<<1];
	void pushUp(int now){
		if(tree[now].lazy > 0)
			tree[now].sum = (msp[tree[now].r+1]-msp[tree[now].l]);
		else if(tree[now].l == tree[now].r)
			tree[now].sum = 0;
		else
			tree[now].sum = tree[now<<1].sum+tree[now<<1|1].sum;
	}
	void build(int now,int l,int r){
		tree[now].l = l;
		tree[now].r = r;
		tree[now].lazy = 0;
		if(l == r){
			tree[now].sum = 0;
			return;
		}
		int Mid = (l+r)>>1;
		build(now<<1,l,Mid);
		build(now<<1|1,Mid+1,r);
		pushUp(now);
	}
	void update(int now,int l,int r,int add){
		if(l <= tree[now].l && tree[now].r <= r){
			tree[now].lazy += add;
			pushUp(now);
			return;
		}
		int Mid = (tree[now].l+tree[now].r)>>1;
		if(l <= Mid)
			update(now<<1,l,r,add);
		if(Mid < r)
			update(now<<1|1,l,r,add);
		pushUp(now);
	}
}sgt;
int main(){
	double x1,x2,y1,y2;
	int l,r,cas = 0;
	while(~scanf("%d",&n) && n){
		for(int i = 0; i < n; i++){
			scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
			line[i*2] = Line(x1,x2,y1,1);
			line[i*2+1] = Line(x1,x2,y2,-1);
			msp[i*2] = x1;
			msp[i*2+1] = x2;
		}
		msp[n*2] = (double)(-INF);
		sort(msp,msp+n*2+1);
		mlen = unique(msp,msp+n*2+1)-msp;
		sort(line,line+n*2);
		ans = 0;
		sgt.build(1,1,mlen);
		for(int i = 0; i < 2*n-1; i++){
			int l = find(line[i].lx);
			int r = find(line[i].rx)-1;
			sgt.update(1,l,r,line[i].cove);
			ans += sgt.tree[1].sum * (line[i+1].h-line[i].h);
		}
		printf("Test case #%d\n",++cas);
		printf("Total explored area: %.2lf\n\n",ans);
	}
	return 0;
}