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=40100;
 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]=1;
   for(int i=0;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]=0;
}
int main()
{
    int cas,a,b,val;
    cin>>cas;
   while(cas--){
     cin>>n>>m;
     cnt=0;
    for(int i=1;i<=n;i++)
        map[i].clear();
   for(int i=1;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=0;i<m;i++)
   {
        ans=0x3f3f3f3f;
         scanf("%d%d",&a,&b);
         dfs(a,b);
         printf("%d
",ans);
   }
    }
 return 0;
}