poj 1436 Horizontally Visible Segments

Horizontally Visible Segments

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 5717 Accepted: 2083

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

150 4 40 3 13 4 20 2 20 2 3

Sample Output

1

/*题意：有很多垂直的线段，两个线段之间存在连线且和其他的线段没有交点为可见，给出一组数据，求三条线段两两可见的组数。题解：将输入的的线段按x从小到大排列，然后先查询再更新，就是每次插入线段之前，先查询即将插入的那条线段和以前插入的线段是否可见，每条线段都有一种颜色，所以用一个vis[i][j]表示i与j可见最后直接暴力枚举就可以过了*/#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int MAXN=16100;bool vis[MAXN>>1][MAXN>>1];struct node{    int l,r;    int color;}tree[MAXN<<2];struct Node{    int x,y1,y2;}p[MAXN>>1];bool cmp(Node a,Node b){    return a.x<b.x;}void build(int l,int r,int node){    tree[node].l=l;    tree[node].r=r;    tree[node].color=0;    if(l==r)        return;    int mid=(l+r)>>1;    build(l,mid,node<<1);    build(mid+1,r,node<<1|1);}void pushdown(int node){    if(!tree[node].color) return;    tree[node<<1].color=tree[node<<1|1].color=tree[node].color;    tree[node].color=0;///取消标记}void query(int l,int r,int node,int val){    if(tree[node].color!=0){        vis[tree[node].color][val]=1;        return;    }    if(tree[node].l==tree[node].r)        return;    pushdown(node);    int mid=(tree[node].l+tree[node].r)>>1;    if(l<=mid)        query(l,r,node<<1,val);    if(r>mid)        query(l,r,node<<1|1,val);}void update(int l,int r,int node,int val){    if(tree[node].l>=l&&tree[node].r<=r){        tree[node].color=val;        return;    }    pushdown(node);    int mid=(tree[node].l+tree[node].r)>>1;    if(l<=mid)        update(l,r,node<<1,val);    if(r>mid)        update(l,r,node<<1|1,val);}int main(){    int t,n,i,j,k;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        int maxn = 0;        for(i=1;i<=n;i++){            scanf("%d%d%d",&p[i].y1,&p[i].y2,&p[i].x);            p[i].y1=2*p[i].y1;            p[i].y2=2*p[i].y2;            maxn = max(maxn,p[i].y1);            maxn = max(maxn,p[i].y2);        }        build(0,maxn,1);        sort(p+1,p+n+1,cmp);        memset(vis,0,sizeof(vis));        for(i=1;i<=n;i++){            query(p[i].y1,p[i].y2,1,i);            update(p[i].y1,p[i].y2,1,i);        }        int ans=0;        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            if(vis[i][j])            for(k=1;k<=n;k++){                if(vis[i][k]&&vis[j][k])                    ans++;            }        printf("%d\n",ans);    }    return 0;}