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poj1436 Horizontally Visible Segments
2015-11-21 01:01:02 来源: 作者: 【 】 浏览:2
Tags:poj1436 Horizontally Visible Segments

Description

There is a number of disjoint vertical line segments in the plane. We say that two segments are horizontally visible if they can be connected by a horizontal line segment that does not have any common points with other vertical segments. Three different vertical segments are said to form a triangle of segments if each two of them are horizontally visible. How many triangles can be found in a given set of vertical segments?


Task

Write a program which for each data set:

reads the description of a set of vertical segments,

computes the number of triangles in this set,

writes the result.

Input

The first line of the input contains exactly one positive integer d equal to the number of data sets, 1 <= d <= 20. The data sets follow.

The first line of each data set contains exactly one integer n, 1 <= n <= 8 000, equal to the number of vertical line segments.

Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:

yi', yi'', xi - y-coordinate of the beginning of a segment, y-coordinate of its end and its x-coordinate, respectively. The coordinates satisfy 0 <= yi' < yi'' <= 8 000, 0 <= xi <= 8 000. The segments are disjoint.

Output

The output should consist of exactly d lines, one line for each data set. Line i should contain exactly one integer equal to the number of triangles in the i-th data set.

Sample Input

1
5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3

Sample Output

1

题意是如果两条线段之间能被一条平行于x轴的线段相连且这条线段和其他线段没有交点,那么这两条线段可见,如果三条线段每两条线段可见,那么他们能组成特定三角形,那么问三角形有多少个。这题先把所有线段储存起来,按x大小升序排列,然后相当于依次读入不同颜色的线段,每次操作,先判断这条线段所在的纵坐标范围内颜色种类,这些颜色种类对应的线段和当前这条线段是可见的,接着把这条线段插入区间,更新总区间的颜色。这里有一点要注意,为了避免单位元线段被“忽略”,把所有的纵坐标都乘2.如3 0 4 1 0 2 2 3 4 2这组数据不乘2的话2-3会被忽略。刚开始所有颜色都为0,如果线段是纯色,那么为大于0的数,若为-1,则是杂色,要在子区间找。

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#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        #include
        #include
        
          #include
         
           #include
          
            #include
           
             using namespace std; struct node{ int l,r,cnt; }b[8*8005]; struct edge{ int y2,y3,x; }s[8005]; bool cmp(edge a,edge b){ return a.x
            
             mid)update(l,r,value,i*2+1); else { update(l,mid,value,i*2); update(mid+1,r,value,i*2+1); } } void question(int l,int r,int id,int i) { int mid; if(b[i].cnt>0){ mark[b[i].cnt][id]=true;return; } if(b[i].cnt==0 || (b[i].l==b[i].r))return; mid=(b[i].l+b[i].r)/2; if(r<=mid)question(l,r,id,i*2); else if(l>mid)question(l,r,id,i*2+1); else { question(l,mid,id,i*2); question(mid+1,r,id,i*2+1); } } int main() { int n,m,i,j,T,x,y2,y3,ans; scanf("%d",&T); while(T--) { scanf("%d",&n); for(i=1;i<=n;i++){ scanf("%d%d%d",&y2,&y3,&x); s[i].y2=2*y2;s[i].y3=2*y3;s[i].x=x; } sort(s+1,s+n+1,cmp); memset(mark,false,sizeof(mark)); build(0,16000,1); for(i=1;i<=n;i++){ question(s[i].y2,s[i].y3,i,1); update(s[i].y2,s[i].y3,i,1); } ans=0; for(i=1;i<=n;i++){ for(j=i+1;j<=n;j++){ if(mark[i][j]) { for(k=j+1;k<=n;k++){ if(mark[j][k] && mark[i][k]){ ans++; //printf("%d %d %d\n",i,j,k); } } } } } printf("%d\n",ans); } return 0; }
            
           
          
         
        
      
     
    
   
  


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