hdu 1542（线段树+扫描线 求矩形相交面积）

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12059    Accepted Submission(s): 5083

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
Source
Mid-Central European Regional Contest 2000
看了大神的解释 懂了一点 http://m.blog.csdn.net/article/details?id=52048323
  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<map>
  5 #include<cstdlib>
  6 #include<vector>
  7 #include<set>
  8 #include<queue>
  9 #include<cstring>
 10 #include<string.h>
 11 #include<algorithm>
 12 #define INF 0x3f3f3f3f
 13 typedef long long ll;
 14 typedef unsigned long long LL;
 15 using namespace std;
 16 const int N=1000+100;
 17 double x[2*N];
 18 struct Edge{
 19     double l,r;
 20     double h;
 21     int flag;
 22 }edge[2*N];
 23 struct node{
 24     int l,r;
 25     int s;
 26     double len;
 27 }tree[N*8];
 28 bool cmp(Edge x,Edge y){
 29     return x.h<y.h;
 30 }
 31 void pushup(int pos){
 32     if(tree[pos].s)tree[pos].len=x[tree[pos].r+1]-x[tree[pos].l];
 33     else if(tree[pos].l==tree[pos].r)tree[pos].len=0;
 34     else{
 35         tree[pos].len=tree[pos<<1].len+tree[pos<<1|1].len;
 36     }
 37 }
 38 void build(int l,int r,int pos){
 39     tree[pos].l=l;tree[pos].r=r;
 40     tree[pos].s=0;tree[pos].len=0;
 41     if(tree[pos].l==tree[pos].r)return;
 42     int mid=(tree[pos].l+tree[pos].r)>>1;
 43     build(l,mid,pos<<1);
 44     build(mid+1,r,pos<<1|1);
 45 }
 46 void update(int l,int r,int pos,int xx){
 47     if(tree[pos].l==l&&tree[pos].r==r){
 48         tree[pos].s=tree[pos].s+xx;
 49         pushup(pos);
 50         return;
 51     }
 52     int mid=(tree[pos].l+tree[pos].r)>>1;
 53     if(r<=mid)update(l,r,pos<<1,xx);
 54     else if(l>mid)update(l,r,pos<<1|1,xx);
 55     else{
 56         update(l,mid,pos<<1,xx);
 57         update(mid+1,r,pos<<1|1,xx);
 58     }
 59     pushup(pos);
 60 }
 61 int main(){
 62     int n;
 63     int tt=0;
 64     while(scanf("%d",&n)!=EOF){
 65         tt++;
 66         if(n==0)break;
 67         int t=0;
 68         for(int i=0;i<n;i++){
 69             double x1,x2,y1,y2;
 70             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
 71             Edge &t1=edge[t];Edge &t2=edge[t+1];
 72             t1.l=t2.l=x1;t1.r=t2.r=x2;
 73             t1.h=y1;t2.h=y2;
 74             t1.flag=1;t2.flag=-1;
 75             x[t]=x1;x[t+1]=x2;
 76             t=t+2;
 77         }
 78         sort(edge,t+edge,cmp);
 79         sort(x,x+t);
 80         int k=1;
 81         for(int i=1;i<t;i++){
 82             if(x[i]!=x[i-1]){
 83                 x[k++]=x[i];
 84             }
 85         }
 86         //cout<<1<<endl;
 87         build(0,k-1,1);
 88         //cout<<2<<endl;
 89         double ans=0.0;
 90         //cout<<t<<endl;
 91         for(int i=0;i<t;i++){
 92             int l=lower_bound(x,x+k,edge[i].l)-x;
 93             int r=lower_bound(x,x+k,edge[i].r)-x-1;
 94             update(l,r,1,edge[i].flag);
 95             ans=ans+(edge[i+1].h-edge[i].h)*tree[1].len;
 96             //cout<<ans<<endl;
 97         }
 98         printf("Test case #%d\n",tt);
 99         printf("Total explored area: %.2f\n\n",ans);
100         //cout<<ans<<endl;
101     }
102 }

 

posted on 2017-03-05 19:19  见字如面  阅读(1018)  评论(0)  编辑  收藏  举报

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