poj 1436 Horizontally Visible Segments(线段树 区间的覆盖关系)

Horizontally Visible Segments

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 1942 Accepted: 736

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

题目：http://poj.org/problem?id=1436

题意：给你n条垂直的线段，这些线段不会相交，如果两条线段直接存在一条水平线可以连接到这两条线段，并且不与其他线段相交，我们说着两条线段可见，求一共有多少对三条线段两两可见。。。

分析：这题当然可以用从上往下的扫描线，加上个排序什么的，来解决，而这题的线段树写法，依旧是模拟区间的覆盖，而这题要求覆盖的关系，一开始我以为在更新线段树的时候就可以判断覆盖关系，然后直接覆盖，最后在写一个遍历整棵树，找出所有覆盖关系，然而这样是错的，因为可能之前那个区间还没找到它盖上的区间就被替换掉了。。。

而正确的做法就是每次先寻找这个区间覆盖了那些线段，然后把它加到线段树里面，这样就避免了之前的错误。。。现在就得到了线段是否可见的关系图，然后暴力枚举三条线段。。。。貌似深搜没法做的样子

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;const int mm=16666;struct seg{    int l,r,x;}g[mm];int head[mm],flag[mm],ver[mm<<8],next[mm<<8];int dly[mm<<2];int edge,ans;bool cmp(seg a,seg b){    return a.x<b.x;}void prepare(){    edge=0;    memset(head,-1,sizeof(head));    memset(dly,0,sizeof(dly));    memset(flag,0,sizeof(flag));}void addedge(int u,int v){    ver[edge]=v,next[edge]=head[u],head[u]=edge++;}void pushdown(int rt){    dly[rt<<1]=dly[rt<<1|1]=dly[rt];    dly[rt]=0;}void updata(int L,int R,int id,int l,int r,int rt){    if(L<=l&&R>=r)    {        dly[rt]=id;        return;    }    if(dly[rt])pushdown(rt);    int m=(l+r)>>1;    if(L<=m)updata(L,R,id,lson);    if(R>m)updata(L,R,id,rson);}void query(int L,int R,int id,int l,int r,int rt){    if(dly[rt])    {        if(flag[dly[rt]]!=id)            addedge(dly[rt],flag[dly[rt]]=id);        return;    }    if(l==r)return;    int m=(l+r)>>1;    if(L<=m)query(L,R,id,lson);    if(R>m)query(L,R,id,rson);}int main(){    int i,j,k,v,n,cs;    scanf("%d",&cs);    while(cs--)    {        scanf("%d",&n);        for(i=0;i<n;++i)        {            scanf("%d%d%d",&g[i].l,&g[i].r,&g[i].x);            g[i].l<<=1,g[i].r<<=1;        }        sort(g,g+n,cmp);        prepare();        for(i=0;i<n;++i)            query(g[i].l,g[i].r,i+1,0,mm,1),            updata(g[i].l,g[i].r,i+1,0,mm,1);        ans=0;        for(i=1;i<=n;++i)            for(j=head[i];j>=0;j=next[j])                for(k=next[j];k>=0;k=next[k])                {                    for(v=head[ver[j]];v>=0;v=next[v])                        if(ver[v]==ver[k])break;                    if(v<0)                    for(v=head[ver[k]];v>=0;v=next[v])                        if(ver[v]==ver[j])break;                    if(v>=0)++ans;                }        printf("%d\n",ans);    }    return 0;}