POJ 1330 Nearest Common ancesters(LCA,Tarjan离线算法）

Description:

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

纯模版题，测试模版，留着用。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<vector>
using namespace std;

const int MAXN=10010;//树中结点的数目

int F[MAXN];//并查集
int r[MAXN];//并查集中集合的个数
bool vis[MAXN];//访问标记
int ancestor[MAXN];//祖先
struct Node
{
int to, next;
}edge[MAXN*2];

int head[MAXN];
int tol;
void addedge(int a,int b)
{
edge[tol].to=b;
edge[tol].next=head[a];
head[a]=tol++;
edge[tol].to=a;
edge[tol].next=head[b];
head[b]=tol++;
}

struct Query
{
int q,next;
int index;//查询编号
}query[MAXN*2];//查询数
int ans[MAXN*2];//查询结果
int cnt;
int h[MAXN];
int tt;
int Q;//查询个数

void add_query(int a,int b,int i)//边a 到 b，第 i 次查询
{
query[tt].q=b;
query[tt].next=h[a];
query[tt].index=i;
h[a]=tt++;
query[tt].q=a;
query[tt].next=h[b];
query[tt].index=i;
h[b]=tt++;
}

void init(int n)//传入n为结点总数
{
for(int i=1;i<=n;i++)
{
F[i]=-1;
r[i]=1;
vis[i]=false;
ancestor[i]=0;
tol=0;
tt=0;
cnt=0;//已经查询到的个数
}
memset(head,-1,sizeof(head));
memset(h,-1,sizeof(h));
}
int find(int x)
{
if(F[x]==-1)return x;
return F[x]=find(F[x]);
}

void Union(int x,int y)//合并
{
int t1=find(x);
int t2=find(y);
if(t1!=t2)
{
if(r[t1]<=r[t2])
{
F[t1]=t2;
r[t2]+=r[t1];
}
else
{
F[t2]=t1;
r[t1]+=r[t2];
}
}
}

void LCA(int u)
{
//if(cnt>=Q)return;//不要加这个
ancestor[u]=u;
vis[u]=true;//这个一定要放在前面
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(vis[v])continue;
LCA(v);
Union(u,v);
ancestor[find(u)]=u;
}
for(int i=h[u];i!=-1;i=query[i].next)
{
int v=query[i].q;
if(vis[v])
{
ans[query[i].index]=ancestor[find(v)];
cnt++;//已经找到的答案数
}
}
}
bool flag[MAXN];
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int N;
memset(flag, true, sizeof(flag));
scanf("%d", &N);
init(N);
int u, v;
for(int i=1;i<N;i++)
{
scanf("%d%d", &u, &v);
flag[v] = false;
addedge(u, v);
}
Q = 1;
scanf("%d%d", &u, &v);
add_query(u, v, 1);
int root;
for(int i=1;i<=N;i++) if(flag[i])
root = i;
LCA(root);
for(int i=1;i<=Q;i++)
printf("%d\n", ans[i]);
}
return 0;
}