poj1330 Nearest Common Ancestors LCA离线tarjan

http://poj.org/problem?id=1330

Nearest Common Ancestors

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18970		Accepted: 10044

Description

A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: 

 
In the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if node x is in the path between the root and node y. For example, node 4 is an ancestor of node 16. Node 10 is also an ancestor of node 16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7. A node x is called a common ancestor of two different nodes y and z if node x is an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is called the nearest common ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y and z among their common ancestors. Hence, the nearest common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is. 

For other examples, the nearest common ancestor of nodes 2 and 3 is node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor of z, then the nearest common ancestor of y and z is y. 

Write a program that finds the nearest common ancestor of two distinct nodes in a tree. 

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,..., N. Each of the next N -1 lines contains a pair of integers that represent an edge --the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers whose nearest common ancestor is to be computed.

Output

Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.

Sample Input

2
16
1 14
8 5
10 16
5 9
4 6
8 4
4 10
1 13
6 15
10 11
6 7
10 2
16 3
8 1
16 12
16 7
5
2 3
3 4
3 1
1 5
3 5

Sample Output

4
3

Source

Taejon 2002

裸的LCA,就当测试模板。。。之前一直没过原因竟然是询问的head数组开小了23333

用的tarjan离线，关于求最近公共祖先的tarjan离线算法，看了这里就会了:http://scturtle.is-programmer.com/posts/30055.html

代码:

/**
 * @author neko01
 */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const int N=10005;
struct edge{
    int to,next;
}e1[N*2];
struct query{
    int to,next,id;
}e2[10];
int h1[N],h2[N];
int ans[10];
int f[N],fa[N];
bool vis[N];
int in[N];
int tot1,tot2;
int n;
void init()
{
    tot1=tot2=0;
    for(int i=0;i<=n;i++)
        h1[i]=-1,h2[i]=-1,vis[i]=false,f[i]=i,in[i]=0;
}
void add1(int u,int v)
{
    e1[tot1].to=v;
    e1[tot1].next=h1[u];
    h1[u]=tot1++;
}
void add2(int u,int v,int id)
{
    e2[tot2].to=v;
    e2[tot2].next=h2[u];
    e2[tot2].id=id;
    h2[u]=tot2++;
}
int find(int x)
{
    return f[x]==x?x:f[x]=find(f[x]);
}
void tarjan(int u)
{
    vis[u]=true;
    for(int i=h1[u];i!=-1;i=e1[i].next)
    {
        int v=e1[i].to;
        if(!vis[v])
        {
            fa[v]=u;
            tarjan(v);
            f[find(v)]=u;
        }
    }
    for(int i=h2[u];i!=-1;i=e2[i].next)
    {
        int v=e2[i].to;
        if(vis[v])
            ans[e2[i].id]=find(v);
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t;
    int u,v;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        init();
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            in[v]++;
            add1(u,v);
            add1(v,u);
        }
        scanf("%d%d",&u,&v);
        add2(u,v,1);
        add2(v,u,1);
        for(int i=1;i<=N;i++)
			if(!in[i])
			{
				tarjan(i);
				break;
			}
        printf("%d\n",ans[1]);
    }
    return 0;
}