poj 1436 Horizontally Visible Segments(线段树 区间的覆盖关系)
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Horizontally Visible Segments
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
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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|1
using 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;
}