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  　　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 
#include 
#include 
#include 
using namespace std;
vector<int> v[41000];
vector<int> w[41000];
int f[41000][21];//f[i][j]表示i点向上2^j层的祖先 
int g[41000][21];//g[i][j]表示i点到从i向上2^j层的祖先的距离 
int dep[41000];
int n,m;
void dfs(int pos,int pre,int depth)
{
    dep[pos]=depth;
    for(int i=0;i)
    {
        int t=v[pos][i];
        if(t==pre) continue;
        f[t][0]=pos;
        g[t][0]=w[pos][i];
        dfs(t,pos,depth+1);
    }
}
int query(int a,int b)
{
    int sum=0;
    if(dep[a]//深度较深的点 
    for(int i=20;i>=0;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=20;i>=0;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][0]+g[b][0];//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,-1,sizeof dep);//多组数据我们初始化 
        memset(f,0,sizeof f);
        memset(g,0,sizeof g);
        for(int i=0;i//md
            v[i].clear(),w[i].clear();
        for(int i=1;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[1].size();
        dfs(1,0,1);//dfs处理出每个点的深度，以及各种... 
    
        for(int i=1;1<)
            for(int j=1;j<=n;j++)
                f[j][i]=f[f[j][i-1]][i-1],
                g[j][i]=g[f[j][i-1]][i-1]+g[j][i-1];
        for(int i=1;i<=m;i++)
        {
            int x,y;
            cin>>x>>y;
            if(x==y) cout<<"0"<<endl;
            else cout<endl;
        }
    }
    return 0;
}