SCU2016-02 A题（LCA）

分析：

求树上最短距离，Tarjan算法处理出距离根节点的距离，然后两个相加减去二倍距离根节点的距离即可。

#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define pr(x) cout << #x << ": " << x << "  " 
#define pl(x) cout << #x << ": " << x << endl;

using namespace std;  
const int N = 50005;  

vector<int> v[N],w[N],query[N],num[N];  
int pre[N],dist[N],ans[N];  
bool vis[N];  
int n;  

void Init()  
{  
    for(int i=1; i<=n; i++)  
    {  
        v[i].clear();  
        w[i].clear();  
        query[i].clear();  
        num[i].clear();  
        pre[i] = i;  
        dist[i] = 0;  
        vis[i] = false;  
    }  
}  

int Find(int x)  
{  
    if(pre[x] != x)  
        pre[x] = Find(pre[x]);  
    return pre[x];  
}  

void Union(int x,int y)  
{  
    x = Find(x);  
    y = Find(y);  
    if(x == y) return;  
    pre[y] = x;  
}  

void Tarjan(int cur,int val)  
{  
    vis[cur] = true;  
    dist[cur] = val;  
    int size = v[cur].size();  
    for(int i=0;i<size;i++)  
    {  
        int tmp = v[cur][i];  
        if(vis[tmp]) continue;  
        Tarjan(tmp,val + w[cur][i]);  
        Union(cur,tmp);  
    }  
    int Size = query[cur].size();  
    for(int i=0;i<Size;i++)  
    {  
        int tmp = query[cur][i];  
        if(!vis[tmp]) continue;  
        ans[num[cur][i]] = dist[cur] + dist[tmp] - 2*dist[Find(tmp)];  
    }  
}  

int main()  
{  
    int T,Q,x,y,z;  
    scanf("%d",&T);  
    while(T--)  
    {  
        scanf("%d%d",&n,&Q);  
        Init();  
        for(int i=1;i<n;i++)  
        {  
            scanf("%d%d%d",&x,&y,&z);  
            v[x].push_back(y);  
            w[x].push_back(z);  
            v[y].push_back(x);  
            w[y].push_back(z);  
        }  
        for(int i=0;i<Q;i++)  
        {  
            scanf("%d%d",&x,&y);  
            query[x].push_back(y);  
            query[y].push_back(x);  
            num[x].push_back(i);  
            num[y].push_back(i);  
        }  
        Tarjan(1,0);  
        for(int i=0;i<Q;i++)  
           printf("%d\n",ans[i]);  
    }  
    return 0;  
}