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

150 4 40 3 13 4 20 2 20 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。

#include<iostream>#include<stdio.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<queue>#include<stack>#include<string>#include<algorithm>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<b.x;}bool mark[8015][8015];int k;void build(int l,int r,int i){int mid;b[i].l=l;b[i].r=r;b[i].cnt=0;if(l==r)return;mid=(l+r)/2;build(l,mid,i*2);build(mid+1,r,i*2+1);}void update(int l,int r,int value,int i){int mid;if(b[i].l==l && b[i].r==r){b[i].cnt=value;return;}if(b[i].cnt!=-1){b[i*2].cnt=b[i*2+1].cnt=b[i].cnt;b[i].cnt=-1;}mid=(b[i].l+b[i].r)/2;if(r<=mid)update(l,r,value,i*2);else if(l>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;}