poj1151 Atlantis

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstdlib>
  4 #include<cmath>
  5 #include<algorithm>
  6 #include<cstring>
  7 using namespace std;
  8 const int N=210;
  9 struct node{
 10     int left,right,c; //c : 区间被覆盖的层数, m: 区间的测度 
 11     double m;
 12 }tree[N*4];
 13 struct Line{
 14     double x,y1,y2;  //纵方向直线, x:直线横坐标, y1 y2:直线上的下面与上面的两个纵坐标
 15     int s;   //s = 1 : 直线为矩形的左边, s = 0:直线为矩形的右边
 16 }line[N];
 17 bool cmp(Line a,Line b){
 18     return a.x<b.x;
 19 }
 20 double ty[N],y[N];   //y[] 整数与浮点数的对应数组; ty[]:用来求y[]的辅助数组
 21 void build(int t,int left,int right){
 22     tree[t].left=left;
 23     tree[t].right=right;
 24     tree[t].c=0;
 25     tree[t].m=0;
 26     if(left+1<right){
 27         int mid=(left+right)>>1;
 28         build(t<<1,left,mid);
 29         build((t<<1)+1,mid,right);
 30     }
 31 }
 32 void update(int t){
 33     if(tree[t].c>0) tree[t].m=y[tree[t].right]-y[tree[t].left];
 34     //将线段树上区间的端点分别映射到 y[]数组所对应的浮点数上,由此计算出测度 
 35     else if(tree[t].left+1==tree[t].right) tree[t].m=0; 
 36     else tree[t].m=tree[t<<1].m+tree[(t<<1)+1].m;
 37 }
 38 void insert(int t,int left,int right){
 39     if(left<=tree[t].left && tree[t].right<=right){
 40         tree[t].c++;
 41         update(t);
 42         return ;
 43     }
 44     int mid=(tree[t].left+tree[t].right)>>1;
 45     if(left<mid) insert(t<<1,left,right);
 46     if(right>mid) insert((t<<1)+1,left,right);
 47     update(t);
 48 }
 49 void del(int t,int left,int right){
 50     if(left<=tree[t].left && tree[t].right<=right){
 51         tree[t].c--;
 52         update(t);
 53         return ;
 54     }
 55     int mid=(tree[t].left+tree[t].right)>>1;
 56     if(left<mid) del(t<<1,left,right);
 57     if(right>mid) del((t<<1)+1,left,right);
 58     update(t);
 59 }
 60 
 61 int getindex(int n,double x){//二分查找出浮点数 t 在数组y[]中的位置(此即所谓的映射关系)
 62     int left,right,mid;
 63     left=1;right=n;
 64     while(left<=right){
 65         mid=(left+right)>>1;
 66         if(y[mid]<x) left=mid+1;
 67         else right=mid-1;
 68     }
 69     return left;
 70 }
 71 int main(){
 72     int n,i;
 73     double x1,y1,x2,y2;
 74     int cas=1;
 75     while(scanf("%d",&n)!=EOF){
 76         if(n==0) break;
 77         for(i=0;i<n;i++){
 78             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
 79             line[i*2].x=x1; line[i*2].y1=y1; line[i*2].y2=y2; line[i*2].s=1;
 80             line[i*2+1].x=x2; line[i*2+1].y1=y1; line[i*2+1].y2=y2; line[i*2+1].s=0;
 81             ty[i*2]=y1;ty[i*2+1]=y2;
 82         }
 83         n<<=1;
 84         sort(line,line+n,cmp);
 85         sort(ty,ty+n);  //默认升序排序
 86         int num=1;
 87         y[1]=ty[0];
 88         for(i=1;i<n;i++){ //离散化处理数组 ty[]使之不含重覆元素,得到新的数组存放到数组y[]中 
 89             if(ty[i]!=ty[i-1]) y[++num]=ty[i];
 90         }
 91         build(1,1,num);//树的叶子节点与数组 y[]中的元素个数相同,以便建立一一对应的关系
 92         int left,right;
 93         double ans=0;
 94         for(i=0;i<n-1;i++){
 95             left=getindex(num,line[i].y1);  //由对应关系计算出线段两端在树中的位置
 96             right=getindex(num,line[i].y2);
 97             if(line[i].s==1)   //插入矩形的左边
 98                 insert(1,left,right);
 99             else         //删除矩形的右边 
100                 del(1,left,right);
101             ans+=tree[1].m*(line[i+1].x-line[i].x);
102         }
103         printf("Test case #%d\n",cas++);
104         printf("Total explored area: %.2lf\n\n",ans);
105     }
106     return 0;
107 }