题干：

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

 

解题报告：

   扫描线模板，裸题。tree[].f代表完全覆盖的次数。注意到，区间更新没有pushdown，是因为我们最终只是使用了tree[1].val这样一个变量。

AC代码：

#include<bits/stdc++.h>

using namespace std;
const double eps = 1e-8;
const int MAX = 100 + 5;
double pre[MAX << 2];
//空间需要开8倍。。
int n;
struct Node {
	double l,r,h;
	int f;
	Node(){}
	Node(double l,double r,double h,int f):l(l),r(r),h(h),f(f){}
	
} seg[16*MAX];
struct TREE {
	int l,r;
	double val;
	int f;
} tree[16*MAX];
bool cmp(const Node & a,const Node & b) {
	return a.h < b.h;
}
void pushup(int cur) {
	if(tree[cur].f >0) tree[cur].val = pre[tree[cur].r+1] - pre[tree[cur].l];
	else {
		if(tree[cur].l == tree[cur].r) tree[cur].val = 0;
		else tree[cur].val=tree[2*cur].val+tree[2*cur+1].val;
	}
}  
void build(int l,int r,int cur) {
	tree[cur].l = l;
	tree[cur].r = r;
	tree[cur].val = 0;
	tree[cur].f = 0;
	if(l == r) return ;
	int m = (l + r)/2;
	build(l,m,2*cur);
	build(m+1,r,2*cur+1); 
}
void scan(int pl,int pr,int f,int cur) {
	if(pl <= tree[cur].l && pr>=tree[cur].r) {
		tree[cur].f+=f;
		pushup(cur);
		return;
	}
	if(pl<=tree[2*cur].r) scan(pl,pr,f,2*cur);
	if(pr>=tree[2*cur+1].l) scan(pl,pr,f,2*cur+1);
	pushup(cur);
}
int main()
{
	int iCase = 0;
	double x1,x2,y1,y2,ans;
	while(~scanf("%d",&n) ) {
		if(n == 0) break;
		//离散化
		int top = 0;
		for(int i = 1; i<=n; i++) {
			scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
			seg[++top] = Node(x1,x2,y1,1); 
			pre[top] = x1;
			seg[++top] = Node(x1,x2,y2,-1); 
			pre[top] = x2;
		} 
		sort(pre+1,pre+2*n+1);
		int x = unique(pre+1,pre+2*n+1) - pre-1;
//		for(int i = 1; i<=x; i++) printf("%f ",pre[i]);
//		printf("\n"); 
		build(1,x,1);
		sort(seg+1,seg+2*n+1,cmp);
		ans = 0;
		for(int i = 1; i<2*n; i++) {
			//左闭右开区间
			int ll = lower_bound(pre+1,pre+x+1,seg[i].l) - (pre+1);
			int rr = lower_bound(pre+1,pre+x+1,seg[i].r) - pre-2;
			ll++,rr++;
		//	printf("ll = %d rr = %d\n",ll,rr);
			scan(ll,rr,seg[i].f,1);
			ans += (seg[i+1].h - seg[i].h) * tree[1].val;
		//	printf("ans = %lf  h2=%f h1=%f val = %f\n",ans,seg[i+1].h,seg[i].h,tree[1].val);
			
		}
		printf("Test case #%d\n",++iCase);
		printf("Total explored area: %.2f\n\n",ans);
		
	}
	
	return 0 ;
}