POJ 1436 区间染色

Horizontally Visible Segments

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4507		Accepted: 1662

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

 

 

题目意思：

给n条垂直x轴的线段，若两个线段之间存在没有其他线段挡着的地方，则称两个线段为可见的。若3条线段两两互为可见，称为一组，求n条线段中有多少组。

 

 

思路：

很明显线段树，按x坐标排序，以y建线段树，每加入一条边就和之前的颜色用visited标记起来，然后暴力三重循环即可（虽然一重循环是8000，但是实际上没这么多）。

注意，插入边的时候，边的两端点*2再插入，还是边界问题。

 

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std;

#define N 10005
#define ll root<<1
#define rr root<<1|1
#define mid (a[root].l+a[root].r)/2


int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
int abs(int x,int y){return x<0?-x:x;}

struct Line{
    int y1, y2, x;
}line[N];

struct node{
    int l, r, val;
    bool f;
}a[N*8];

int n;
bool visited[8002][8002];

bool cmp(Line a,Line b){
    return a.x<b.x;
}

void build(int l,int r,int root){
    a[root].l=l;
    a[root].r=r;
    a[root].val=0;
    a[root].f=false;
    if(l==r) return;
    build(l,mid,ll);
    build(mid+1,r,rr);
}

void down(int root){
    if(a[root].f&&a[root].val>0&&a[root].l!=a[root].r){
        a[ll].val=a[rr].val=a[root].val;
        a[root].val=-1;
        a[ll].f=a[rr].f=true;
    }
}
void update(int l,int r,int val,int root){
    //if(!a[root].f) a[root].f=true;
    if(a[root].val==val) return;
    if(a[root].l==l&&a[root].r==r){
        if(!a[root].f){
            a[root].f=true;
            a[root].val=val;
            return;
        }
        else{
            if(a[root].val>0){
                if(!visited[a[root].val][val]){
                    visited[a[root].val][val]=visited[val][a[root].val]=true;
                }
                a[root].val=val;
                return;
            }
        }
    }
    down(root);
    if(r<=a[ll].r) update(l,r,val,ll);
    else if(l>=a[rr].l) update(l,r,val,rr);
    else{
        update(l,mid,val,ll);
        update(mid+1,r,val,rr);
    }
    if(a[ll].f||a[rr].f) a[root].f=true;
    if(a[ll].val==a[rr].val&&a[ll].val>0) a[root].val=a[ll].val;
}

void out(int root){
    if(a[root].l==a[root].r) {
        printf("%d ",a[root].val);return;
    }
    down(root);
    out(ll);
    out(rr);
}
main()
{
    int t, i, j, k;
    cin>>t;
    while(t--){
        scanf("%d",&n);
        int minh=999999999, maxh=-1;
        for(i=0;i<n;i++){
            scanf("%d %d %d",&line[i].y1,&line[i].y2,&line[i].x);
            minh=min(min(line[i].y1,line[i].y2),minh);
            maxh=max(max(line[i].y1,line[i].y2),maxh);
        } 
        build(minh*2,maxh*2,1);
        sort(line,line+n,cmp);
        memset(visited,false,sizeof(visited));
        for(i=0;i<n;i++) update(line[i].y1*2,line[i].y2*2,i+1,1);//,out(1),cout<<endl;
        int ans=0;
        
        for(i=1;i<=n;i++){
            for(j=i+1;j<=n;j++){
                if(visited[i][j]){
                    for(k=j+1;k<=n;k++){
                        if(visited[j][k]&&visited[i][k]){
                            ans++;
                        }
                    }
                }
            }
        }
        printf("%d
",ans);
        
    }
}