HDU 2586 How far away ？

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6309    Accepted Submission(s): 2368

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

23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output

1025100100

 

题目的意思就是给定一些houses的编号，及这些编号房子之间的距离，然后有m次询问，每次询问给出两个房子的编号，要求给出这两个房子的最小距离。这张图是树图，也就是总共N个点N-1条边，然后能把所有点全部连通。本身分析题目很容易知道本题就是求最小公共祖先的问题，首先确定一个根节点，然后DFS遍历一遍计算此根节点到所有节点的距离，然后可以用离线的tarjan算法，来找到两个询问节点a和b的最近公共祖先c，然后要求的结果就是dist[a] - dist[c] + dist[b] - dist[c]，意思很好理解，画棵树看看就可以了，而且这种思路也是很容易想到的。这里用了一下非递归的DFS直接来求，每次询问直接使用DFS求亮点之间的距离，也可以很轻松的AC。

#include <cstdio>  #include <vector>  #include <stack>  #include <string>    using namespace std;    const int N = 40005;    int n, m, t;  vector<int> adj[N];  vector<int> wei[N];  int dist[N];  bool visit[N];    void init(int n)  {      for (int i = 0; i <= n; ++i)      {          adj[i].clear();          wei[i].clear();      }  }    int dfs(int x, int y)  {      int res = 0;        memset(visit, 0, sizeof(visit));      memset(dist, 0, sizeof(dist));        stack<int> st;      st.push(x);      visit[x] = true;        while (!st.empty())      {          int tx = st.top();          st.pop();            if (tx == y)break;            for (int i = 0; i < adj[tx].size(); ++i)          {              int ty = adj[tx][i];              if (visit[ty])continue;              st.push(ty);              visit[ty] = true;                dist[ty] = dist[tx] + wei[tx][i];          }      }      return dist[y];  }    int main()  {      int a, b, w;        scanf("%d", &t);        while (t--)      {          scanf("%d %d", &n, &m);            init(n);            for (int i = 0; i < n - 1; ++i)          {              scanf("%d %d %d", &a, &b, &w);              adj[a].push_back(b);              adj[b].push_back(a);              wei[a].push_back(w);              wei[b].push_back(w);          }            for (int i = 0; i < m; ++i)          {              scanf("%d %d", &a, &b);              printf("%d\n", dfs(a, b));          }                    if (t != 0)printf("\n");      }      return 0;  }