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：

 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 node

10 {

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);  //ÎÞÏòͼ

49       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 }

View Code