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

题目大意：给出n条垂直于x轴的线段的数据y1,y2,x，求出有几个三条线段一组的三元组并且他们兩兩能相见的。
思路：对y轴建树，将x排序，然后按顺序边询问边擦入，用mark[i][j]表示i往左可以看到j。最后用一个三重循环计算答案。
但是注意:0,4,1 和 0,2,2 和 3,4,2这三条线段覆盖的结果是区间0~4通过线段树查找可见线段是两条,其实是3条(2~3可见另一条)
        这里可以将y轴×2表示。这样就能解决这样的问题

/*
 * Author:  Joshua
 * Created Time:  2014年07月19日 星期六 15时26分20秒
 * File Name: poj1436.cpp
 */
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define maxn 8008
#define L(x) (x <<1)
#define R(x) (x << 1 |1)

int T,maxv;
bool mark[maxn][maxn];
struct node
{
    int l,r,color;
} e[maxn << 4];
struct edge
{
    int y1,y2,x;
    edge(int Y1=0,int Y2=0,int X=0):y1(Y1),y2(Y2),x(X){}
    bool operator < (const edge &a) const 
    {
        return x<a.x;
    }
} s[maxn];
void build(int x,int l,int r)
{
    e[x].color=0;
    e[x].l=l;
    e[x].r=r;
    if (l==r) return;
    int mid = l+r >>1;
    build(L(x),l,mid);
    build(R(x),mid+1,r);
}
void query(int x,int l,int r,int v)
{
   if (e[x].color)
   {
       mark[v][e[x].color]=true;
       return;
   }
   int left=e[x].l,right=e[x].r;
   if (left==right) return;
   int mid = left+right >>1;
   if (l<=mid) query(L(x),l,r,v);
   if (r>mid) query(R(x),l,r,v);
}

void pushDown(int x)
{
    if (e[x].color)
    {
        e[L(x)].color=e[R(x)].color=e[x].color;
        e[x].color=0;
    }
}

void updata(int x,int l,int r,int v)
{
    int left=e[x].l,right=e[x].r;
    if (l<=left && right<=r)
    {
        e[x].color=v;
        return;
    }
    pushDown(x);
    int mid=left +right >>1;
    if (l<=mid) updata(L(x),l,r,v);
    if (r>mid) updata(R(x),l,r,v);
}
   
void solve()
{
     int n,x,y1,y2;
     scanf("%d",&n);
     maxv=0;
     for (int i=1;i<=n;++i)
     {
         scanf("%d%d%d",&y1,&y2,&x);
         y1 <<=1;
         y2 <<=1;
         s[i]=edge(y1,y2,x);
         if (y2>maxv) maxv=y2;
     }
     build(1,0,maxv);
     sort(s+1,s+n+1);
     memset(mark,0,sizeof(mark));
     for (int i=1;i<=n;++i)
     {
         query(1,s[i].y1,s[i].y2,i);
         updata(1,s[i].y1,s[i].y2,i);
     }
     int ans=0;
     for (int i=1;i<=n;++i)
         for (int j=i+1;j<=n;++j)
             if (mark[j][i])
                 for (int k=1;k<=i;++k)
                     if (mark[i][k] && mark[j][k])
                         ++ans;
     printf("%d\n",ans);
}
         
int main()
{
    scanf("%d",&T);
    while (T)
    {
        solve();
        --T;
    }

    return 0;
}