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。

#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
",i,j,k);
					  }
				    }
				}
			}
		}
		printf("%d
",ans);
	}
	return 0;
}