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
编程是一种快乐，享受代码带给我的乐趣！！！
