Atlantis

题目连接

http://poj.org/problem?id=1151

Description

here 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

Sample Input

          2 
         
10 10 20 20 
         
15 15 25 25.5 
         
0

Sample Output

Test case #1
Total explored area: 180.00

HINT

题意

给你N个矩形，求矩形相交的面积

题解：

线段树，扫描线模板题

具体请看这一篇博文：http://blog.csdn.net/shiqi_614/article/details/6821814

代码：

  1 //#include<bits/stdc++.h>
  2 #include<cstdio>
  3 #include<iostream>
  4 #include<algorithm>
  5 using namespace std;
  6 #define N 10050
  7 int n,cnt,tot,cas;
  8 double kth[N];
  9 struct Query
 10 {
 11   double l,r,h; int id;
 12   bool operator <(const Query&b)const
 13   {return h<b.h;}
 14 }que[N<<1];
 15 struct Tree{int l,r,lazy;double sum;}tr[N<<2];
 16 template<typename T>void read(T&x)
 17 {
 18   int k=0;char c=getchar();
 19   x=0;
 20   while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
 21   if(c==EOF)exit(0);
 22   while(isdigit(c))x=x*10+c-'0',c=getchar();
 23   x=k?-x:x;
 24 }
 25 void push_up(int x)
 26 {
 27   double len=(kth[tr[x].r]-kth[tr[x].l]);
 28   tr[x].sum=0;
 29   if (tr[x].lazy>0)tr[x].sum=len;
 30   else if(tr[x].r-tr[x].l>1)tr[x].sum=tr[x<<1].sum+tr[x<<1|1].sum;
 31 }
 32 void bt(int x,int l,int r)
 33 {
 34   tr[x]=Tree{l,r,0,0};
 35   if(r-l==1)return;
 36   int mid=(l+r)>>1;
 37   bt(x<<1,l,mid);
 38   bt(x<<1|1,mid,r);
 39 }
 40 void update(int x,int l,int r,int tt)
 41 {
 42   if (l<=tr[x].l&&tr[x].r<=r)
 43     {
 44       tr[x].lazy+=tt;
 45       push_up(x);
 46       return;
 47     }
 48   int mid=(tr[x].l+tr[x].r)>>1;
 49   if(l<mid)update(x<<1,l,r,tt);
 50   if(mid<r)update(x<<1|1,l,r,tt);
 51   push_up(x);
 52 }
 53 double query(int x,int l,int r)
 54 {
 55   if (l<=tr[x].l&&tr[x].r<=r) return tr[x].sum;
 56   int mid=(tr[x].l+tr[x].r)>>1;
 57   double ans=0;
 58   if(l<mid)ans+=query(x<<1,l,r);
 59   if(mid<r)ans+=query(x<<1|1,l,r);
 60   return ans;
 61 }
 62 void input()
 63 {
 64   read(n);
 65   if (n==0)exit(0);
 66   double x1,y1,x2,y2;
 67   for(int i=1;i<=n;i++)
 68     {
 69       //read(x1);read(y1); read(x2);read(y2);
 70       scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
 71       que[++tot]=Query{x1,x2,y1,1};
 72       que[++tot]=Query{x1,x2,y2,-1};
 73       kth[++cnt]=x1;kth[++cnt]=y1;
 74       kth[++cnt]=x2;kth[++cnt]=y2;
 75     }
 76 }
 77 void work()
 78 {
 79   double ans=0;
 80   sort(que+1,que+tot+1);
 81   sort(kth+1,kth+cnt+1);
 82   cnt=unique(kth+1,kth+cnt+1)-kth-1;
 83   bt(1,1,cnt);
 84   for(int i=1;i<=tot-1;i++)
 85     {
 86       int l=lower_bound(kth+1,kth+cnt+1,que[i].l)-kth;
 87       int r=lower_bound(kth+1,kth+cnt+1,que[i].r)-kth;
 88       update(1,l,r,que[i].id);
 89       ans+=tr[1].sum*(que[i+1].h-que[i].h);
 90     }
 91   //printf("Test case #%d\nTotal explored area: %.2lf\n\n",++cas,ans);
 92   printf("Test case #%d\n", ++cas);
 93   printf("Total explored area: %.2f\n\n", ans);
 94 }
 95 void clear(){cnt=0;tot=0;}
 96 int main()
 97 {
 98   #ifndef ONLINE_JUDGE
 99   freopen("aa.in","r",stdin);
100   #endif
101   while(1)
102     {
103       clear();
104       input();
105       work();
106     }
107 }

POJ 1151 Atlantis 矩形面积求交/线段树扫描线