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 }

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