HDU - 2586 How far away ？ (Lca)

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

题解：裸的lca。。。。。。。。

萌新的第一道lca

#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<map>
#include<queue>
#include<iostream>
#include<algorithm>
#include<set>

using namespace std;

const int maxn=100100;

int dp[maxn][30],ff[maxn],vs[maxn];

struct dd
{
    int n,e,v;
};dd d[maxn];

int h[maxn],top,dk[maxn],f[maxn];

struct no{
    int i,v;
    bool operator <(const no &a)const{
        return v>a.v;
    }
};

void s0(int s,int e,int v){
    d[top].e=e;
    d[top].v=v;
    d[top].n=h[s];
    h[s]=top++;
}
priority_queue < no >q;

void gdk(int s){
    dk[s]=0;
    q.push((no){s,0});
    while(!q.empty()){
        no p=q.top();
        q.pop();
        if(f[p.i]==-1)f[p.i]=1;else continue;
        for(int i=h[p.i];i!=-1;i=d[i].n){
            int v=d[i].v,y=d[i].e;
            if(dk[y]>dk[p.i]+v){
                dk[y]=dk[p.i]+v;
                q.push((no){y,dk[y]});
            }
        }
    }
}
int top1,in[maxn];
void dfs(int fath,int i,int k){

    vs[++top1]=i;
    ff[top1]=k;
    in[i]=top1;
    for(int j=h[i];j!=-1;j=d[j].n){
        if(fath==d[j].e)continue;
        dfs(i,d[j].e,k+1);
        vs[++top1]=i;
        ff[top1]=k;
    }
}
int gmi(int x,int y){
    return ff[x]<ff[y]?x:y;
}
void st(int n)
{
    for(int i=1; i<=n; i++)dp[i][0]=i;
    for(int j=1; (1<<j)<=n; j++)
    {
        for(int i=1; i+(1<<j)-1<=n; i++)
        {
            dp[i][j]=gmi(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}
int rmq(int l,int r)
{
    int k=0;
    if(l>r)swap(l,r);
    while((1<<(k+1))<=r-l+1)k++;
    return gmi(dp[l][k],dp[r-(1<<k)+1][k]);
}
void init(int m){
    memset(h,-1,sizeof(h));
    memset(f,-1,sizeof(f));
    top1=0,top=0;
    for(int i=1;i<=m;i++)dk[i]=999999999;
}
int main()
{
    int m,n;
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&m,&n);
        init(m);
        int s,e,v;
        for(int i=0;i<m-1;i++){
            scanf("%d%d%d",&s,&e,&v);
            s0(s,e,v);
            s0(e,s,v);
        }
        gdk(1);
        dfs(-1,1,0);
        st(top1);
        while(n--){
            scanf("%d%d",&s,&e);
            int fa=vs[rmq(in[s],in[e])];
            printf("%d\n",dk[e]+dk[s]-2*dk[fa]);
        }
    }
    return 0;
}