How far away ？

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

Total Submission(s): 2561 Accepted Submission(s): 946

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 经典问题：求最近公共祖先。

//Accepted	2586	125MS	4056K	1713 B	C++#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             using namespace std;const int maxn = 40100;int f[maxn];int d[maxn]; //保存每个节点的深度。vector
             
               a[maxn];map
              
                h; //保存每个节点到其父亲边的距离。int n, m;void getDep(int num, int dep) { d[num] = dep; vector
               
                ::iterator it; it = a[num].begin(); for(; it < a[num].end(); ++it) { getDep(*it, dep+1); }}int work(int a, int b) { int s1 = 0; int s2 = 0; while(a!=b) { while(d[a]
                
                 d[b]) { s2 += h[a]; a = f[a]; } if(d[a]==d[b]&&a!=b) { s1 += h[b]; b = f[b]; } } return s1+s2;}void init() { for(int i = 0; i < n; i++) { a[i].clear(); } h.clear();}int main(){ int T; int u; scanf("%d", &T); int from, to, w; for(u = 0; u < T; u++) { scanf("%d%d", &n, &m); for(int i = 0; i < n-1; i++) { scanf("%d%d%d", &from, &to, &w); a[from].push_back(to); f[to] = from; h[to] = w;//保存每个节点到其父亲的距离 } getDep(1, 0); //query for(int i = 0; i < m; i++) { scanf("%d%d", &from, &to); int res = work(from, to); cout << res << endl; } init(); } return 0;}

转载于:https://www.cnblogs.com/xinyuyuanm/archive/2013/04/18/3029497.html

你可能感兴趣的文章