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<?-x:x;}

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

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

 int n;
 bool visited[][];

 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=;
     a[root].f=false;
     if(l==r) return;
     build(l,mid,ll);
     build(mid+,r,rr);
 }

 void down(int root){
     if(a[root].f&&a[root].val>&&a[root].l!=a[root].r){
         a[ll].val=a[rr].val=a[root].val;
         a[root].val=-;
         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>){
                 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+,r,val,rr);
     }
     if(a[ll].f||a[rr].f) a[root].f=true;
     if(a[ll].val==a[rr].val&&a[ll].val>) 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=, maxh=-;
         for(i=;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*,maxh*,);
         sort(line,line+n,cmp);
         memset(visited,false,sizeof(visited));
         for(i=;i<n;i++) update(line[i].y1*,line[i].y2*,i+,);//,out(1),cout<<endl;
         int ans=;

         for(i=;i<=n;i++){
             for(j=i+;j<=n;j++){
                 if(visited[i][j]){
                     for(k=j+;k<=n;k++){
                         if(visited[j][k]&&visited[i][k]){
                             ans++;
                         }
                     }
                 }
             }
         }
         printf("%d\n",ans);

     }
 }