HDU 2586.How far away ？-离线LCA(Tarjan)

2586.How far away ？

这个题以前写过在线LCA(ST)的，HDU2586.How far away ？-在线LCA(ST)

现在贴一个离线Tarjan版的

代码:

//A-HDU2586-LCA-tarjan离线版
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;

const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=4e4+10;
const int maxm=100+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

struct node{
    int x,y;
    //node(int xx,int yy){
    //    x=xx,y=yy;
    //}node(){}
};

vector<node> edge[maxn],q[maxn];
int ans[maxn],dis[maxn],fa[maxn],vis[maxn];
int n,m,aa,bb,cc;

int find(int x)//找父节点(祖先)
{
    return x==fa[x]?x:fa[x]=find(fa[x]);
}

void unio(int x,int y)//并查集压缩路径
{
    fa[find(y)]=find(x);
}

void init()//初始化
{
    for(int i=1;i<=n;i++){
        edge[i].clear();
        q[i].clear();
        fa[i]=i;//初始化自己的父亲节点是自己
        vis[i]=0;
        ans[i]=0;
        dis[i]=0;
    }
}

void dfs(int x)//tarjan
{
    vis[x]=1;
    for(int i=0;i<edge[x].size();i++){
        int v=edge[x][i].x;
        if(!vis[v]){
            dis[v]=dis[x]+edge[x][i].y;
            dfs(v);
            unio(x,v);
        }
    }
    for(int i=0;i<q[x].size();i++){
        int v=q[x][i].x;
        if(vis[v])
            ans[q[x][i].y]=dis[x]+dis[v]-2*dis[find(v)];
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<n;i++){
            scanf("%d%d%d",&aa,&bb,&cc);
            //edge[aa].push_back(node(bb,cc));
            //edge[bb].push_back(node(aa,cc));
            edge[aa].push_back({bb,cc});
            edge[bb].push_back({aa,cc});
        }
        for(int i=1;i<=m;i++){
            scanf("%d%d",&aa,&bb);
            //q[aa].push_back(node(bb,i));
            //q[bb].push_back(node(aa,i));
            q[aa].push_back({bb,i});
            q[bb].push_back({aa,i});

        }
        dfs(1);
        for(int i=1;i<=m;i++)
        printf("%d
",ans[i]);
    }
}

溜了。