poj 1436 Horizontally Visible Segments(线段树成段覆盖问题+简单hash),好题，覆盖问题想法较难

1、http://poj.org/problem?id=1436

2、题目大意：

给出n段垂直的线段，如果两条线段可以用一条水平的线连接起来，且中间不能跟别的垂直的线相交，那么这两条线段称作是可见的，如果有三条线段两两是可见的，那么称作是线段三角形，求给出的这n条垂直的线段可以组成多少个线段三角形？

3、思路分析

要求多少个线段三角形，就是求有多少个三条线段两两相交的，首先我们先求出两条线段相交的情况，也就是求一下跟第一条线段相交的线段有那几条，定义一个vector容器，比如跟第一条相交的有2,3,4；跟第二条线段相交的有3

那么vector里存的就是这样的：

1   2,3,4

2   3

那么我们知道了他们之间的关系，就可以三重for循环遍历一遍求出来即可

例如我们现在已经知道1-2,1-3，我们要证明1,2,3是线段三角形，只需再证明2-3也是可见的就行了，此时去2后面查找是否有3,如果有那么这个就是线段三角形，计数器+1,

对于区间覆盖来说，用一个cover[rt]数组表示，例如cover[rt]=1,表示rt对应的区间被第一条边覆盖了

4、题目：

Horizontally Visible Segments

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 3463 Accepted: 1275

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

Source

Central Europe 2001

5、AC代码：

#include<stdio.h>#include<string.h>#include<vector>#include<algorithm>using namespace std;#define N 16005#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1int cover[N*4];int hash[N*4];vector<int> adj[N];struct node{    int s;    int e;    int x;}a[N];void pushdown(int rt){    if(cover[rt]!=-1)    {        cover[rt<<1]=cover[rt<<1|1]=cover[rt];        cover[rt]=-1;    }}void query(int L,int R,int id,int l,int r,int rt){    if(cover[rt]!=-1)    {        if(hash[cover[rt]]!=id)        {            adj[cover[rt]].push_back(id);            hash[cover[rt]]=id;        }        return ;    }    if(l==r)    return ;    pushdown(rt);    int m=(l+r)>>1;    if(L<=m)    query(L,R,id,lson);    if(R>m)    query(L,R,id,rson);}void update(int L,int R,int id,int l,int r,int rt){    if(L<=l && R>=r)    {        cover[rt]=id;        return ;    }    pushdown(rt);    int m=(l+r)>>1;    if(L<=m)    update(L,R,id,lson);    if(R>m)    update(L,R,id,rson);}int cmp(node a,node b){    return a.x<b.x;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d%d%d",&a[i].s,&a[i].e,&a[i].x);            a[i].s<<=1;            a[i].e<<=1;            adj[i].clear();//注意清空        }        sort(a,a+n,cmp);        memset(cover,-1,sizeof(cover));        memset(hash,-1,sizeof(hash));        for(int i=0;i<n;i++)        {            query(a[i].s,a[i].e,i,0,N,1);            update(a[i].s,a[i].e,i,0,N,1);        }        for(int i=0;i<n;i++)        {            //这样就可以给vector容器中每一行从小到大排序            sort(adj[i].begin(),adj[i].end());        }        //printf("*************\n");//        for(int i=0;i<n;i++)//        {//            printf("%d:",i);//            for(int j=0;j<adj[i].size();j++)//            printf("%d ",adj[i][j]);//            printf("\n");//        }       // printf("******************\n");        int ans=0;        for(int i=0;i<n;i++)        {            int len=adj[i].size();            for(int j=0;j<len;j++)            {                for(int k=j+1;k<len;k++)                {                    int p=adj[i][j];                    int q=adj[i][k];                    //printf("%d %d\n",p,q);                    if(binary_search(adj[p].begin(),adj[p].end(),q))                    {                        ans++;                    }                }            }        }        printf("%d\n",ans);    }    return 0;}