题目连接

  

How far away ？

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

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

tarjan离线求LCA。。

#include 
     
      using namespace std;const int N = 40010;struct Tarjan_Lca {    bool vis[N];    vector
      
       
        > A, Q[N];    int tot, par[N], ans[210], head[N], dist[N];    struct edge { int to, w, next; }G[N << 1];    inline void init(int n) {        A.clear();        for(int i = 0; i < n + 2; i++) {            par[i] = i;            vis[i] = false;            head[i] = -1;            dist[i] = 0;            Q[i].clear();        }    }    inline void add_edge(int u, int v, int w) {        G[tot] = { v, w, head[u] }, head[u] = tot++;        G[tot] = { u, w, head[v] }, head[v] = tot++;    }    inline void built(int n, int m) {        int u, v, w;        while(n-- > 1) {            scanf("%d %d %d", &u, &v, &w);            add_edge(u, v, w);        }        for(int i = 0; i < m; i++) {            scanf("%d %d", &u, &v);            A.push_back(pair
        
         (u, v));            Q[u].push_back(pair
         
          (v, i));            Q[v].push_back(pair
          
           (u, i)); } } inline int find(int x) { while(x != par[x]) { x = par[x] = par[par[x]]; } return x; } inline void tarjan(int u, int fa) { for(int i = head[u]; ~i; i = G[i].next) { edge &e = G[i]; if(e.to == fa) continue; dist[e.to] = dist[u] + e.w; tarjan(e.to, u); vis[e.to] = true; par[e.to] = u; } for(auto &r: Q[u]) { if(vis[r.first]) ans[r.second] = find(r.first); } } inline void solve(int n, int m) { init(n); built(n, m); tarjan(1, 1); for(int i = 0; i < m; i++) { printf("%d\n", dist[A[i].first] + dist[A[i].second] - 2 * dist[ans[i]]); } }}go;int main() {#ifdef LOCAL freopen("in.txt", "r", stdin); freopen("out.txt", "w+", stdout);#endif int t, n, m; scanf("%d", &t); while(t--) { scanf("%d %d", &n, &m); go.solve(n, m); } return 0;}