hdu----（2586）How far away ？（DFS/LCA/RMQ）

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

ECJTU 2009 Spring Contest

 

Recommend

 

用邻接表+dfs比较容易过...

代码：

 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 #include<vector>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 const int maxn=;
  struct node
 {
   int id,val;
 };
 bool vis[maxn];
 vector< node >map[maxn];
 node tem;
 int n,m,ans,cnt;
 void dfs(int a,int b)
 {

    if(a==b){
        if(ans>cnt)ans=cnt;
        return ;
    }
    int Size=map[a].size();
       vis[a]=;
    for(int i=;i<Size;i++){
     if(!vis[map[a][i].id]){
      cnt+=map[a][i].val;
      dfs(map[a][i].id,b);
      cnt-=map[a][i].val;
     }
    }
    vis[a]=;
 }
 int main()
 {
     int cas,a,b,val;
     cin>>cas;
    while(cas--){
      cin>>n>>m;
      cnt=;
     for(int i=;i<=n;i++)
         map[i].clear();
    for(int i=;i<n;i++){
       scanf("%d%d%d",&a,&b,&val);

       tem=(node){b,val};
       map[a].push_back(tem);  //ÎÞÏòͼ
       tem=(node){a,val};
       map[b].push_back(tem);
    }
    for(int i=;i<m;i++)
    {
         ans=0x3f3f3f3f;
          scanf("%d%d",&a,&b);
          dfs(a,b);
          printf("%d\n",ans);
    }
     }
  return ;
 }