How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5492    Accepted Submission(s): 2090

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

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

 

 

Source

 

 

Recommend

 

用邻接表+dfs比较容易过...

代码：

1 #include
       
         2 #include
        
          3 #include
         
           4 #include
          
            5 #include
           
             6 #include
            
              7 using namespace std; 8 const int maxn=40100; 9 struct node10 {11 int id,val;12 };13 bool vis[maxn];14 vector< node >map[maxn];15 node tem;16 int n,m,ans,cnt;17 void dfs(int a,int b)18 {19 20 if(a==b){21 if(ans>cnt)ans=cnt;22 return ;23 }24 int Size=map[a].size();25 vis[a]=1;26 for(int i=0;i
             
              >cas;39 while(cas--){40 cin>>n>>m;41 cnt=0;42 for(int i=1;i<=n;i++)43 map[i].clear();44 for(int i=1;i