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

How far away ？

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21250 Accepted Submission(s): 8368

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

3 2

1 2 10

3 1 15

1 2

2 3

2 2

1 2 100

1 2

2 1

Sample Output

100

题目大意：给一个有权树，有Q次询问，求任意两点的距离。

题目分析：在树中有一个性质：任意两点的距离 ANS = dist[ u ] + dist[ v ] - dist[ lca(u,v) ]，其中dist[ I ]是根到 I 的距离，由于需要多次询问，所以使用Tarjan离线求LCA比较快

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<queue>
  5 #include<vector>
  6 using namespace std;
  7 const int maxn=40005;
  8 struct edge{
  9     int to;
 10     int len;
 11 };
 12 vector<struct edge>G[maxn];
 13 vector<int>qury[maxn];
 14 vector<int>num[maxn];
 15 int dis[maxn],vis[maxn],ans[maxn],fa[maxn];
 16 void init()
 17 {
 18     memset(dis,0,sizeof(dis));
 19     memset(vis,0,sizeof(vis));
 20     for(int i = 0 ; i < 40005 ; i++)
 21     {
 22         G[i].clear();
 23         qury[i].clear();
 24         num[i].clear();
 25         fa[i]=i;
 26     }
 27 }
 28 int find(int x)
 29 {
 30     int xx=x;
 31     while(fa[x]!=x)
 32     {
 33         x=fa[x];
 34     }
 35     while(fa[xx]!=x)
 36     {
 37         int t=fa[xx];
 38         fa[xx]=x;
 39         xx=t;
 40     }
 41     return x;
 42 }
 43 void Union(int x,int y)
 44 {
 45     int xx=find(x);
 46     int yy=find(y);
 47     if(xx!=yy);
 48     fa[yy]=xx;//在完成子节点的Tarjan遍历之后，把子节点纳入父节点名下
 49 }
 50 void Tarjan(int u,int ll)
 51 {
 52     vis[u]=1;
 53     dis[u]=ll;
 54     for(int i = 0 ; i< G[u].size() ;i++)
 55     {
 56         struct edge wqw=G[u][i];
 57         if(vis[wqw.to])continue;
 58         Tarjan(wqw.to,wqw.len+ll);
 59         Union(u,wqw.to);
 60     }
 61     for(int i = 0 ; i < qury[u].size() ; i++)
 62     {
 63         if(vis[qury[u][i]])
 64         {
 65             ans[num[u][i]]=dis[u]+dis[qury[u][i]]-2*dis[find(qury[u][i])];    
 66         }
 67     }
 68 }
 69 int main()
 70 {
 71     int t;
 72     scanf("%d",&t);
 73     while(t--)
 74     {
 75         init();
 76         int n,m;
 77         scanf("%d%d",&n,&m);
 78         for(int i = 1 ; i < n ; i++)
 79         {
 80             int a,b,c;
 81             scanf("%d%d%d",&a,&b,&c);
 82             G[a].push_back((struct edge){b,c});
 83             G[b].push_back((struct edge){a,c});
 84         }
 85         for(int i = 0 ; i < m ; i++)
 86         {
 87             int a,b;
 88             scanf("%d%d",&a,&b);
 89             qury[a].push_back(b);
 90             qury[b].push_back(a);
 91             num[a].push_back(i);
 92             num[b].push_back(i);
 93         }
 94         Tarjan(1,0);
 95         for(int i = 0 ; i < m ; i++)
 96         {
 97             printf("%d\n",ans[i]);
 98         }
 99     }
100     return 0; 
101 }

【HDOJ2586】【Tarjan离线求LCA】

How far away ？

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