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

2024-07-21 07:11:16   230人阅读

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
23 21 2 103 1 151 22 32 21 2 1001 22 1
 

 

Sample Output
1025100100
 

 

Source
ECJTU 2009 Spring Contest
 

 

Recommend
 
用邻接表+dfs比较容易过...
代码：
 1 #include<cstring> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<vector> 5 #include<algorithm> 6 #include<iostream> 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<Size;i++){27     if(!vis[map[a][i].id]){28      cnt+=map[a][i].val;29      dfs(map[a][i].id,b);30      cnt-=map[a][i].val;31     }32    }33    vis[a]=0;34 }35 int main()36 {37     int cas,a,b,val;38     cin>>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<n;i++){45       scanf("%d%d%d",&a,&b,&val);46 47       tem=(node){b,val};48       map[a].push_back(tem);  //ÎÞÏòͼ49       tem=(node){a,val};50       map[b].push_back(tem);51    }52    for(int i=0;i<m;i++)53    {54         ans=0x3f3f3f3f;55          scanf("%d%d",&a,&b);56          dfs(a,b);57          printf("%d\n",ans);58    }59     }60  return 0;61 }
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hdu----（2586）How far away ？（DFS/LCA/RMQ）


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