hdu 2586 LCA dfs+ST求LCA

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

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#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn=50005;
typedef long long ll;
int n,m,cnt,tot;
bool vis[maxn];
int head[maxn],deep[maxn*4],ind[maxn*4],first[maxn],dp[maxn*4][25];
ll dist[maxn];
struct node{
    int to,next;
    ll w;
}e[maxn*3];
inline ll read(){
    char x=getchar();    ll num=0;
    while(!(x>=48&&x<=57))
        x=getchar();
    while(x>=48&&x<=57)
        num=num*10+x-48,x=getchar();
    return num;
}
inline void add(int u,int v,ll w){
    e[cnt].to=v;
    e[cnt].next=head[u];
    e[cnt].w=w;
    head[u]=cnt++;
}
inline void dfs(int t,int dep){
    vis[t]=1;
    ind[++tot]=t;//给每个结点编号
    deep[tot]=dep;
    first[t]=tot;
    for(int i=head[t];~i;i=e[i].next){
        int v=e[i].to; ll w=e[i].w;
        if(!vis[v]){
            dist[v]=dist[t]+w;
            dfs(v,dep+1);
        }
        ind[++tot]=t;
        deep[tot]=dep;
    }
}
inline void ST(int n){
    for(int i=1;i<=n;i++) dp[i][0]=i;//保存最小深度下标
    for(int j=1;(1<<j)<=n;j++){
        for(int i=1;i+(1<<j)-1<=n;i++){
            int a=dp[i][j-1],b=dp[i+(1<<(j-1))][j-1];
            dp[i][j]=deep[a]>deep[b]?b:a;
        }
    }
}
inline int RMQ(int l,int r){
    int k=0;
    while((1<<(k+1))<=r-l+1) k++;
    int a=dp[l][k],b=dp[r-(1<<k)+1][k];
    return deep[a]>deep[b]?b:a;
}
inline int LCA(int u,int v){
    int x=first[u],y=first[v];
    if(x>y) swap(x,y);
    int res=RMQ(x,y);
    return ind[res];
}
inline void init(int n){
    tot=cnt=0;
    memset(head,-1,sizeof(int)*(n+1));
    memset(vis,0,sizeof(vis));
}

int main(){
    int T;
    cin>>T;
    while(T--){
        n=read();int q=read();
        init(n);
        int x,y;
        for(int i=1;i<n;i++){
            x=read();y=read(); ll w=read();
            add(x,y,w);
            add(y,x,w);
        }
        dist[1]=0;
        dfs(1,1);
        ST(tot);
        while(q--){
            int a,b;
            a=read();b=read();
            int lca=LCA(a,b);
            printf("%lld\n",dist[a]+dist[b]-2*dist[lca]);
        }
    }

    return 0;
}