Problem Description

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

 

Sample Output

 

Source

#pragma comment(linker, "/STACK:102400000,102400000")

#include<iostream>

#include<cstring>

#include<cstdio>

#include<vector>

#define MAX_N 40003

using namespace std;

bool vis[MAX_N];

struct edge {

public:

    int to;

    long long cost;

    edge(int t, long long c) : to(t), cost(c) { }

    edge() { }

};

vector<edge> G[MAX_N];

int q,n,m;

struct node {

public:

    int p, v;

    node(int pp, int vv) : p(pp), v(vv) { }

    node() { }

};

vector<node> Q[MAX_N];

int lca[MAX_N], ancestor[MAX_N];

int T, cas = ;

int father[MAX_N];

void init() {

    for (int i = ; i <= n; i++)

        father[i] = i;

}

int Find(int u) {

    if (u == father[u])return u;

    else return father[u] = Find(father[u]);

}

void unionSet(int u,int v) {

    int x = Find(u), y = Find(v);

    if (x == y)return;

    father[x] = y;

}

bool Same(int u,int v) {

    return Find(u) == Find(v);

}

long long dis[MAX_N];

void Tarjan(int u,int p) {

    for (int i = ; i < G[u].size(); i++) {

        int v = G[u][i].to;

        if (v == p)continue;

        dis[v] = dis[u] + G[u][i].cost;

        Tarjan(v, u);

        unionSet(u, v);

        ancestor[Find(u)] = u;

    }

    vis[u] = ;

    for (int i = ; i < Q[u].size(); i++) {

        int v = Q[u][i].v;

        if (vis[v])

            lca[Q[u][i].p] = ancestor[Find(v)];

    }

}

pair<int, int> qu[MAX_N];

int main() {

    //cin.sync_with_stdio(false);

    scanf("%d", &T);

    while (T--) {

        scanf("%d%d", &n, &q);

        m = n - ;

        init();

        memset(vis, , sizeof(vis));

        memset(ancestor, , sizeof(ancestor));

        memset(dis, , sizeof(dis));

        memset(lca, , sizeof(lca));

        for (int i = ; i <= n; i++)G[i].clear();

        for (int i = ; i <= n; i++)Q[i].clear();

        for (int i = ; i < m; i++) {

            int u, v;

            long long c;

            scanf("%d%d%I64d", &u, &v, &c);

            G[u].push_back(edge(v, c));

            G[v].push_back(edge(u, c));

        }

        for (int i = ; i < q; i++) {

            int u, v;

            scanf("%d%d", &u, &v);

            qu[i] = make_pair(u, v);

            Q[u].push_back(node(i, v));

            Q[v].push_back(node(i, u));

        }

        Tarjan(, );

        for (int i = ; i < q; i++)

            printf("%I64d\n", dis[qu[i].first] + dis[qu[i].second] -  * dis[lca[i]]);

    }

    return ;

}