Horizontally Visible Segments

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 6290		Accepted: 2287

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.

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.

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.

1
5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3

1

  1 //线段树区间更新，端点放大2倍
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<cstdlib>
  8 using namespace std;
  9 typedef long long ll;
 10 const int maxn=8e3+10;
 11
 12 #define lson l,m,rt<<1
 13 #define rson m+1,r,rt<<1|1
 14
 15 int col[maxn<<2];
 16 bool mp[maxn][maxn];
 17
 18 struct node{
 19     int y1,y2,x,id;
 20 }line[maxn];
 21
 22 bool cmp(node a,node b)
 23 {
 24     return a.x<b.x;
 25 }
 26
 27 void init()
 28 {
 29     memset(mp,0,sizeof mp);
 30     memset(line,0,sizeof line);
 31     memset(col,0,sizeof col);
 32 }
 33
 34 void pushdown(int rt)
 35 {
 36     if(col[rt]){
 37         col[rt<<1]=col[rt<<1|1]=col[rt];
 38         col[rt]=0;
 39     }
 40 }
 41
 42 void update(int L,int R,int c,int l,int r,int rt)
 43 {
 44     if(L<=l&&r<=R){
 45         col[rt]=c;
 46         return ;
 47     }
 48
 49     pushdown(rt);
 50     int m=(l+r)>>1;
 51     if(L<=m) update(L,R,c,lson);
 52     if(R> m) update(L,R,c,rson);
 53 }
 54
 55 void query(int L,int R,int c,int l,int r,int rt)
 56 {
 57     if(col[rt]){
 58         mp[c][col[rt]]=1;
 59         return ;
 60     }
 61
 62     if(l==r){
 63         return ;
 64     }
 65
 66     int m=(l+r)>>1;
 67     if(L<=m) query(L,R,c,lson);
 68     if(R> m) query(L,R,c,rson);
 69 }
 70
 71 int main()
 72 {
 73     int t;
 74     scanf("%d",&t);
 75     while(t--){
 76         init();
 77         int n;
 78         scanf("%d",&n);
 79         int N=-1;
 80         for(int i=1;i<=n;i++){
 81             scanf("%d%d%d",&line[i].y1,&line[i].y2,&line[i].x);
 82             line[i].id=i;line[i].y1*=2;line[i].y2*=2;N=max(N,max(line[i].y1,line[i].y2));
 83         }
 84         sort(line+1,line+1+n,cmp);//按x轴从小到大排序，降维，只对y轴进行建树
 85         for(int i=1;i<=n;i++){
 86             query(line[i].y1,line[i].y2,line[i].id,0,N,1);
 87             update(line[i].y1,line[i].y2,line[i].id,0,N,1);
 88         }
 89         int ans=0;
 90         for(int i=1;i<=n;i++){
 91             for(int j=1;j<=n;j++){
 92                 if(mp[i][j]){
 93                     for(int k=1;k<=n;k++){
 94                         if(mp[i][k]&&mp[k][j]) ans++;
 95                     }
 96                 }
 97             }
 98         }
 99         printf("%d\n",ans);
100     }
101 }