LCA好（Codeforces Round #294 (Div. 2)E. A and B and Lecture Rooms）

E. A and B and Lecture Rooms
A and B are preparing themselves for programming contests.

The University where A and B study is a set of rooms connected by corridors. Overall, the University has
n rooms connected by
n - 1 corridors so that you can get from any room to any other one by moving along the corridors. The rooms are numbered from
1 to n.

Every day А and B write contests in some rooms of their university, and after each contest they gather together in the same room and discuss problems. A and B want the distance from the rooms where problems are discussed to the rooms where contests are written
to be equal. The distance between two rooms is the number of edges on the shortest path between them.

As they write contests in new rooms every day, they asked you to help them find the number of possible rooms to discuss problems for each of the following
m days.

Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of rooms in the University.

The next n - 1 lines describe the corridors. The
i-th of these lines (1 ≤ i ≤ n - 1) contains two integers
ai and
bi (1 ≤ ai, bi ≤ n), showing that the
i-th corridor connects rooms
ai and
bi.

The next line contains integer m (1 ≤ m ≤ 105) — the number of queries.

Next m lines describe the queries. The
j-th of these lines (1 ≤ j ≤ m) contains two integers
xj and
yj (1 ≤ xj, yj ≤ n) that means that on the
j-th day A will write the contest in the room
xj, B will write in the room
yj.

Output
In the i-th (1 ≤ i ≤ m) line print the number of rooms that are equidistant from the rooms where A and B write contest on the
i-th day.

Sample test(s)

Input

4
1 2
1 3
2 4
1
2 3

Output

1

Input

4
1 2
2 3
2 4
2
1 2
1 3

Output

0
2

Note
in the first sample there is only one room at the same distance from rooms number 2 and 3 — room number 1.

题意：给定一棵树，M次询问，每次询问(u,v)到达这两个点距离相等的点有多少个

思路：求出u，v的公共祖先LCA，然后分情况讨论就可以了

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=100010;
int N,M;
vector<int> edge[maxn];
int dep[maxn],fa[maxn],anc[maxn][30];
int num[maxn];
void dfs(int u,int f,int depth)
{
    dep[u]=depth;
    fa[u]=f;
    num[u]=1;
    int len=edge[u].size();
    for(int i=0;i<len;i++)
    {
        int v=edge[u][i];
        if(v==f)continue;
        dfs(v,u,depth+1);
        num[u]+=num[v];
        fa[v]=u;
    }
}
void preprocess()
{
    for(int i=1;i<=N;i++)
    {
        anc[i][0]=fa[i];
        for(int j=1;(1<<j)<=N;j++)anc[i][j]=-1;
    }
    for(int j=1;(1<<j)<=N;j++)
    {
        for(int i=1;i<=N;i++)
        {
            if(anc[i][j-1]==-1)continue;
            int a=anc[i][j-1];
            anc[i][j]=anc[a][j-1];
        }
    }
}
int LCA(int p,int q)
{
    if(dep[p]<dep[q])swap(p,q);
    int k=0;
    while((1<<(k+1))<=dep[p])k++;
    for(int i=k;i>=0;i--)
        if(dep[p]-(1<<i)>=dep[q])
        p=anc[p][i];
    if(p==q)return p;
    for(int i=k;i>=0;i--)
        if(anc[p][i]!=-1&&anc[p][i]!=anc[q][i])
            p=anc[p][i],q=anc[q][i];
    return fa[p];
}
int jump(int u,int dis)
{
    if(dis==0)return u;
    for(int i=19;i>=0;i--)
        if((1<<i)<=dis)
            u=anc[u][i],dis-=(1<<i);
    return u;
}
int solve(int u,int v)
{
    if(u==v)return N;
    if(dep[u]<dep[v])swap(u,v);
    int k=LCA(u,v);
    int dis=dep[u]+dep[v]-dep[k]*2;
    if(dis&1)return 0;
    dis>>=1;
    int mid=jump(u,dis),ans=0;
    if(k==mid)
    {
        u=jump(u,dis-1);
        v=jump(v,dis-1);
        return N-num[u]-num[v];

    }
    else
    {
        u=jump(u,dis-1);
        return num[mid]-num[u];
    }
}
int main()
{
    while(scanf("%d",&N)!=EOF)
    {
        int u,v;
        for(int i=0;i<=N;i++)edge[i].clear();
        for(int i=1;i<N;i++)
        {
            scanf("%d%d",&u,&v);
            edge[u].push_back(v);
            edge[v].push_back(u);
        }
        dfs(1,-1,0);
        preprocess();
        scanf("%d",&M);
        while(M--)
        {
            scanf("%d%d",&u,&v);
            printf("%d\n",solve(u,v));
        }
    }
    return 0;
}