1、http://poj.org/problem?id=1436
2、题目大意：
给出n段垂直的线段，如果两条线段可以用一条水平的线连接起来，且中间不能跟别的垂直的线相交，那么这两条线段称作是可见的，如果有三条线段两两是可见的，那么称作是线段三角形，求给出的这n条垂直的线段可以组成多少个线段三角形？
3、思路分析
要求多少个线段三角形，就是求有多少个三条线段两两相交的，首先我们先求出两条线段相交的情况，也就是求一下跟第一条线段相交的线段有那几条，定义一个vector容器，比如跟第一条相交的有2,3,4；跟第二条线段相交的有3
那么vector里存的就是这样的：
1 2,3,4
2 3
那么我们知道了他们之间的关系，就可以三重for循环遍历一遍求出来即可
例如我们现在已经知道1-2,1-3，我们要证明1,2,3是线段三角形，只需再证明2-3也是可见的就行了，此时去2后面查找是否有3,如果有那么这个就是线段三角形，计数器+1,
对于区间覆盖来说，用一个cover[rt]数组表示，例如cover[rt]=1,表示rt对应的区间被第一条边覆盖了
4、题目：
Horizontally Visible Segments
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 3463 | Accepted: 1275 |
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.
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.
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
Source
Central Europe 2001
5、AC代码：
#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
#define N 16005
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int cover[N*4];
int hash[N*4];
vector<int> adj[N];
struct node
{int s;int e;int x;
}a[N];
void pushdown(int rt)
{if(cover[rt]!=-1){cover[rt<<1]=cover[rt<<1|1]=cover[rt];cover[rt]=-1;}
}
void query(int L,int R,int id,int l,int r,int rt)
{if(cover[rt]!=-1){if(hash[cover[rt]]!=id){adj[cover[rt]].push_back(id);hash[cover[rt]]=id;}return ;}if(l==r)return ;pushdown(rt);int m=(l+r)>>1;if(L<=m)query(L,R,id,lson);if(R>m)query(L,R,id,rson);}
void update(int L,int R,int id,int l,int r,int rt)
{if(L<=l && R>=r){cover[rt]=id;return ;}pushdown(rt);int m=(l+r)>>1;if(L<=m)update(L,R,id,lson);if(R>m)update(L,R,id,rson);
}
int cmp(node a,node b)
{return a.x<b.x;
}
int main()
{int t,n;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d%d%d",&a[i].s,&a[i].e,&a[i].x);a[i].s<<=1;a[i].e<<=1;adj[i].clear();//注意清空}sort(a,a+n,cmp);memset(cover,-1,sizeof(cover));memset(hash,-1,sizeof(hash));for(int i=0;i<n;i++){query(a[i].s,a[i].e,i,0,N,1);update(a[i].s,a[i].e,i,0,N,1);}for(int i=0;i<n;i++){//这样就可以给vector容器中每一行从小到大排序sort(adj[i].begin(),adj[i].end());}//printf("*************\n");
// for(int i=0;i<n;i++)
// {
// printf("%d:",i);
// for(int j=0;j<adj[i].size();j++)
// printf("%d ",adj[i][j]);
// printf("\n");
// }// printf("******************\n");int ans=0;for(int i=0;i<n;i++){int len=adj[i].size();for(int j=0;j<len;j++){for(int k=j+1;k<len;k++){int p=adj[i][j];int q=adj[i][k];//printf("%d %d\n",p,q);if(binary_search(adj[p].begin(),adj[p].end(),q)){ans++;}}}}printf("%d\n",ans);}return 0;
}