HDU - 2586 How far away ？（暴力 | LCA）

Posted 2020-10-29 leonard-

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2586

题目：

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

 

题意：给定一棵带权树，m次询问，两个点之间的距离。

题解：暴力DFS。

 1 #include <vector>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int N=12345;
 7 int fa[N],ans;
 8 vector < pair<int,int> > E[N];
 9 vector < pair<int,int> > q[N];
10 bool root[N],vis[N];
11 
12 void init(){
13     for(int i=0;i<N;i++){
14         fa[i]=i;
15         E[i].clear();
16         q[i].clear();
17         vis[i]=false;
18         root[i]=true;
19     }
20 }
21 
22 int Find(int x){
23     return fa[x]==x?x:fa[x]=Find(fa[x]);
24 }
25 
26 void Union(int x,int y){
27     int fx=Find(x),fy=Find(y);
28     if(fx!=fy){
29         fa[fy]=fx;
30     }
31 }
32 
33 void solve(int u){
34     vis[u]=1;
35     for(int i=0;i<E[u].size();i++){
36         int v=E[u][i];
37         if(vis[v]) continue;
38         solve(v);
39         Union(u,v);
40     }
41     for(int i=0;i<q[u].size();i++){
42         int v=q[u][i];
43         if(!vis[v]) continue;
44         ans=Find(v);
45     }
46 }
47 
48 int main(){
49     int t,n,m;
50     scanf("%d",&t);
51     while(t--){
52         int u,v,w;
53         init();
54         scanf("%d%d",&n,&m);
55         for(int i=1;i<n;i++){
56             scanf("%d%d%d",&u,&v,&w);
57             E[u].push_back(make_pair(v,w));
58             E[v].push_back(make_pair(u,w));
59         }
60         for(int i=1;i<=m;i++){
61             scanf("%d%d",&u,,&v);
62             Q[u].push_back(make_pair(v,i));
63             Q[v].push_back(make_pair(u,i));
64         }
65         tarjan();
66         
67     }
68     return 0;
69 }