HDU2586 How far away ？（LCA板子题）

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.

InputFirst 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.OutputFor 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
这个题关于log的常数优化貌似不写也行，但最好还是写一下（洛谷那道模板不写的话会T两个点）

#include <bits/stdc++.h>
#define N 50005
using namespace std;
int n,m,t,tot=0,head[N],ver[2*N],edge[2*N],Next[2*N],f[N][22],d[N],dist[N],lg[50005];;
void add(int x,int y,int z)
{
    ver[++tot]=y,edge[tot]=z,Next[tot]=head[x],head[x]=tot;
}
void prework()
{
    queue<int>q;
    q.push(1);
    d[1]=1;
    while(q.size())
    {
        int x=q.front();
        q.pop();
        int i;
        for(i=head[x];i;i=Next[i])
        {
            int y=ver[i],z=edge[i];
            if(d[y])continue;
            d[y]=d[x]+1;
            dist[y]=dist[x]+z;
            f[y][0]=x;//y节点的2^0倍祖先是x
            int j;
            for(j=1;j<=lg[d[x]];j++)
            {
                f[y][j]=f[f[y][j-1]][j-1];
            }
            q.push(y);    
        }
    }
}
int lca(int x,int y)
{
    if(d[x]<d[y])swap(x,y);//使得x深度较小 
    int i;
    while(d[x] > d[y])
    {
        x = f[x][lg[d[x]-d[y]] - 1];
    }
    if(x==y)return x;
    for(i=lg[d[x]] - 1;i>=0;i--)
    {
        if(f[x][i]!=f[y][i])x=f[x][i],y=f[y][i];
    }
    return f[x][0];
}
int main()
{
    int t;
    cin>>t;
    for(int i = 1; i <= 50005; ++i)
        lg[i] = lg[i-1] + (1 << lg[i-1] == i);//常数优化得放到循环外面 
    while(t--)
    {
        cin>>n>>m;
        int i;
        for(i=1;i<=n;i++)d[i]=head[i]=0;
        tot=0;//记得归零！ 
        for(i=1;i<=n-1;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
            add(y,x,z);
        }
        prework();
        for(i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            printf("%d
",dist[a]+dist[b]-2*dist[lca(a,b)]);
        }
    }
    return 0;
}