（线段树）poj1436-Horizontally Visible Segments

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

就是区间染色问题。用线段树维护每个区间的颜色，对于读入数据处理排序为从左到右update，x坐标其实无关紧要。用二维bool数组记录两个线段是否水平可视，枚举即可求得答案。

#include <iostream>
//#include<bits/stdc++.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int MAX=16000;
int d,n;
bool link[8001][8001];
struct node
{
    int col;
}st[20*MAX];
void init(int l,int r,int k)
{
    st[k].col=0;
    if(l==r)
        return;
    init(l,(l+r)/2,2*k);
    init((l+r)/2+1,r,2*k+1);
}
void pushdown(int k)
{
    if(st[k].col==0)
        return;
    st[2*k].col=st[2*k+1].col=st[k].col;
    st[k].col=0;
}
void update(int l,int r,int L,int R,int k,int j)//[l,r]是目标区间 [L,R]是当前区间,j是向左看的线的编号
{
    if(L>=l&&R<=r)
    {
        st[k].col=j;
        return;
    }
    if(L==R)
        return;
    pushdown(k);
    if(l<=(L+R)/2)
        update(l,r,L,(L+R)/2,2*k,j);
    if(r>(L+R)/2)
        update(l,r,(L+R)/2+1,R,2*k+1,j);
}
void query(int l,int r,int L,int R,int k,int j)//[L,R]是当前区间,[l,r]是目标区间
{
    if(st[k].col)
    {
        link[st[k].col][j]=true;
        return;
    }
    if(L==R)
        return;
    if(l<=(L+R)/2)
        query(l,r,L,(L+R)/2,2*k,j);
    if(r>(R+L)/2)
        query(l,r,(L+R)/2+1,R,2*k+1,j);
}
struct re
{
    int x,xia,shang;
}z[MAX];
bool cmp(re aa,re bb)
{
    if(aa.x!=bb.x)
    return aa.x<bb.x;
    else
        return aa.xia<bb.xia;
}
int main()
{
    scanf("%d",&d);
    int x,s,i,tem;
    while(d--)
    {
        scanf("%d",&n);
        init(0,MAX,1);
        memset(link,false,sizeof(link));
//        for(i=1;i<=n;i++)
//        {
//            scanf("%d %d %d",&x[i],&s,&tem);
//            update(2*x,2*s,0,MAX,1,i);
//        }
        for(i=1;i<=n;i++)
            scanf("%d %d %d",&z[i].xia,&z[i].shang,&z[i].x);
        sort(z+1,z+n+1,cmp);
        for(i=1;i<=n;i++)
            {
                query(2*z[i].xia,2*z[i].shang,0,MAX,1,i);
                update(2*z[i].xia,2*z[i].shang,0,MAX,1,i);

            }
        int j,k,q;
        int an=0;
        for(i=1;i<=n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                if(link[i][j])
                {
//                    printf("%d %d i:%d %d %d j:%d %d %d
",i,j,2*z[i].xia,2*z[i].shang,z[i].x,2*z[j].xia,2*z[j].shang,z[j].x);
                    for(q=i+1;q<j;q++)
                    {
                        if(link[i][q]&&link[q][j])
                            an++;
                    }
                }
            }
        }
        printf("%d
",an);
    }
}