题意：求矩形面积的并
思路：
注意是[l,mid][mid,r] 这是真正的线段了
就当扫描线模板使吧~

//By SiriusRen
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define eps 1e-5
int N,tot,n,cases;
double y[666],ans=0;
struct Node{double x,y1,y2;int cover;}node[666];
struct Tree{double l,r,len;int cover;}tree[66666];
bool cmp(Node a,Node b){return a.x<b.x;}
void build(int l,int r,int pos){tree[pos].l=y[l],tree[pos].r=y[r],tree[pos].len=tree[pos].cover=0;if(l+1==r)return;int mid=(l+r)>>1,lson=pos<<1,rson=pos<<1|1;build(l,mid,lson),build(mid,r,rson);
}
void push_up(int l,int r,int pos){if(tree[pos].cover>0)tree[pos].len=tree[pos].r-tree[pos].l;else if(l+1==r)tree[pos].len=0;else tree[pos].len=tree[pos<<1].len+tree[pos<<1|1].len;
}
void update(int l,int r,int pos,Node jy){if(tree[pos].l>=jy.y1&&tree[pos].r<=jy.y2){tree[pos].cover+=jy.cover;push_up(l,r,pos);return;}int mid=(l+r)>>1,lson=pos<<1,rson=pos<<1|1;if(tree[lson].r>=jy.y2)update(l,mid,lson,jy);else if(tree[rson].l<=jy.y1)update(mid,r,rson,jy);else update(l,mid,lson,jy),update(mid,r,rson,jy);push_up(l,r,pos);
}
int main(){while(scanf("%d",&N)&&N){ans=tot=0;for(int i=1;i<=N;i++){double X1,Y1,X2,Y2;scanf("%lf%lf%lf%lf",&X1,&Y1,&X2,&Y2);node[++tot].x=X1,node[tot].y1=Y1,node[tot].y2=Y2;node[tot].cover=1;y[tot]=Y1;node[++tot].x=X2,node[tot].y1=Y1,node[tot].y2=Y2;node[tot].cover=-1;y[tot]=Y2;}sort(node+1,node+1+tot,cmp),sort(y+1,y+1+tot);n=unique(y+1,y+1+tot)-y-1;build(1,n,1),update(1,n,1,node[1]);for(int i=2;i<=tot;i++){ans+=tree[1].len*(node[i].x-node[i-1].x);update(1,n,1,node[i]);}printf("Test case #%d\nTotal explored area: %.2lf\n\n",++cases,ans);}
}