POJ Horizontally Visible Segments

Horizontally Visible Segments

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 2318		Accepted: 885

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

Source

Central Europe 2001

 

题意：垂直线段的水平可见的条件为它们之间可以通过一条水平直线相连，且不经过任何一条垂直线段。如果有三条线段两两可见，那么它们是一个垂直线段的水平可见三角形。给定n条线段，问有多少这样的垂直线段的水平可见三角形。

 

首先 总坐标都得乘以2 ； 因为假设你有一线段[5,6] ,[3,5],[6,7],[5,6];如果不乘以2，第一条和第四条线段是不可见的，但事实是可见的。。。。自己想象下吧

思路：先按坐标x排好序，再依次查询和更新 ，我的更新和查询直接用一个函数带过，f[k].cover = -1 【f[k].left ,f[k].right】之间覆盖有不同的垂直线段，f[k].cover!=-1时，后面查询的线段覆盖前面的。。。line[i][j] = 1表示可见 。。。

最后一个三重循环就能得到答案了

 

#include<stdio.h>
#include<algorithm> 
#include<string.h>
#include<iostream>
using namespace std;
struct SegmentLine{
    int x,y1,y2 ;
}s[8010] ;
bool line[8010][8010] ;
struct SegmentTree{
    int left,right,cover ;
    int mid(){    return (left + right) >> 1 ;}
}f[80000] ;
bool cmp(SegmentLine a,SegmentLine b){
    return a.x < b.x ;
}
void build(int left,int right,int k){
    f[k].left = left  ;
    f[k].right = right ;
    f[k].cover = 0;
    if(left == right)     return ;
    int mid = f[k].mid() ;
    build(left,mid,k<<1) ;
    build(mid+1,right,k<<1|1) ;
}
void query(int left,int right,int idx,int k){
    if(left <= f[k].left && right >= f[k].right && f[k].cover != -1){
        line[f[k].cover][idx] = 1 ;
        f[k].cover = idx ;
        return ;
    }
    if(f[k].cover != -1){
        f[k<<1].cover = f[k<<1|1].cover = f[k].cover ;
        f[k].cover = -1;
    }
    int mid = f[k].mid() ;
    if(right <= mid)
        query(left,right,idx,k<<1) ;
    else    if(left > mid)
        query(left,right,idx,k<<1|1) ;
    else    {
        query(left,mid,idx,k<<1) ;
        query(mid+1,right,idx,k<<1|1) ;
    }
}
int main(){
    int T ;
    scanf("%d",&T) ;
    while(T--){
        int n;
        scanf("%d",&n) ;
        build(0,16000,1) ;
        memset(line,0,sizeof(line)) ;
        for(int i=1;i<=n;i++){
            scanf("%d%d%d",&s[i].y1,&s[i].y2,&s[i].x) ;
            s[i].y1 *= 2 ;
            s[i].y2 *= 2 ;
        }
        sort(&s[1],&s[1]+n,cmp) ;
        for(int i=1;i<=n;i++){
            query(s[i].y1,s[i].y2,i,1) ;
        }
        int ans = 0 ;
        for(int i=1;i<n;i++){
            for(int j=i+1;j<=n;j++){
                if(line[i][j])
                    for(int k=j+1;k<=n;k++)
                        if(line[j][k] && line[i][k])
                            ans++ ;
            }
        }
        printf("%d\n",ans) ;
    }
    return 0;
}