HDU2586How far away ？

http://acm.hdu.edu.cn/showproblem.php?pid=2586

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13821    Accepted Submission(s):
5195

Problem Description

　　There are n houses in the village and some
bidirectional roads connecting them. Every day peole always like to ask like
this "How far is it if I want to go from house A to house B"? Usually it hard to
answer. But luckily int this village the answer is always unique, since the
roads are built in the way that there is a unique simple path("simple" means you
can't visit a place twice) between every two houses. Yout task is to answer all
these curious people.

 

Input

　　First line is a single integer T(T<=10), indicating
the number of test cases.
　　For each test case,in the first line there are
two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses
and the number of queries. The following n-1 lines each consisting three numbers
i,j,k, separated bu a single space, meaning that there is a road connecting
house i and house j,with length k(0<k<=40000).The houses are labeled from
1 to n.
  　　Next m lines each has distinct integers i and j, you areato answer
the distance between house i and house j.

 

Output

　　For each test case,output m lines. Each line represents
the answer of the query. Output a bland line after each test case.

 

Sample Input

　　2

　　3 2

　　1 2 10

　　3 1 15

　　1 2

　　2 3

　　

　　2 2

　　1 2 100

　　1 2

　　2 1

 

Sample Output

　　10
25
100
100

 

Source

ECJTU
2009 Spring Contest

　　题目大意：有T组数据 每组给出n个点由n-1条路连接，给出m次询问，求a和b两个村庄的距离

　　我实在不想吐槽，多组数据不用初始化也能A？中间把最近公共祖先输出了也能A！？逗我玩呢。

　　带领小学妹谢JustPenz233

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
vector<int> v[];
vector<int> w[];
int f[][];//f[i][j]表示i点向上2^j层的祖先
int g[][];//g[i][j]表示i点到从i向上2^j层的祖先的距离
int dep[];
int n,m;
void dfs(int pos,int pre,int depth)
{
    dep[pos]=depth;
    for(int i=;i<v[pos].size();i++)
    {
        int t=v[pos][i];
        if(t==pre) continue;
        f[t][]=pos;
        g[t][]=w[pos][i];
        dfs(t,pos,depth+);
    }
}
int query(int a,int b)
{
    int sum=;
    if(dep[a]<dep[b]) swap(a,b);//深度较深的点
    for(int i=;i>=;i--)//找到a在深度dep[b]处的祖先
    {
        if(dep[f[a][i]]>=dep[b])
        {
            sum+=g[a][i];//a到该祖先的距离
            a=f[a][i];
        }
    }
    if(a==b) return sum;//挪到相同深度后如果在同一点直接return
    int x;
    for(int i=;i>=;i--)//否则a和b一起往上蹦跶
    {
        if(f[a][i]!=f[b][i])
        {
            sum+=g[a][i];
            a=f[a][i];
            sum+=g[b][i];
            b=f[b][i];
        }
    }
    return sum+g[a][]+g[b][];//最后蹦跶到最近公共祖先的下一层，所以要再加上上一层
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        memset(dep,-,sizeof dep);//多组数据我们初始化
        memset(f,,sizeof f);
        memset(g,,sizeof g);
        for(int i=;i<n;i++)//md
            v[i].clear(),w[i].clear();
        for(int i=;i<n;i++)
        {
            int x,y,c;
            cin>>x>>y>>c;
            v[x].push_back(y);
            w[x].push_back(c);
            v[y].push_back(x);
            w[y].push_back(c);
        }
        int xxx=v[].size();
        dfs(,,);//dfs处理出每个点的深度，以及各种... 
 
        for(int i=;<<i<=n;i++)
            for(int j=;j<=n;j++)
                f[j][i]=f[f[j][i-]][i-],
                g[j][i]=g[f[j][i-]][i-]+g[j][i-];
        for(int i=;i<=m;i++)
        {
            int x,y;
            cin>>x>>y;
            if(x==y) cout<<""<<endl;
            else cout<<query(x,y)<<endl;
        }
    }
    return ;
}