• HDU-2586 How far away?


    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.
    InputFirst 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.OutputFor 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
    
    
     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<vector>
     4 #include<algorithm>
     5 #include<string.h>
     6 using namespace std;
     7 int ans[250],vis[40500];
     8 vector< pair <int ,int > > vec[40500];
     9 vector <pair <int ,int > > query[40500];
    10 struct Node{
    11     int f;
    12     int len;
    13 } pre[40500];
    14 Node Find(int x)
    15 {
    16     if(pre[x].f==x)
    17     {
    18         return pre[x];
    19     }
    20     else
    21     {
    22         Node now;
    23         now=Find(pre[x].f);
    24         pre[x].len+=now.len;
    25         pre[x].f=now.f;
    26         return pre[x];
    27     }
    28 }
    29 int dfs(int u,int fa,int length)
    30 {
    31     pre[u].f=u;
    32     pre[u].len=0;
    33     vis[u]=1;
    34     for(int i=0;i<vec[u].size();i++)
    35     {
    36         int v=vec[u][i].first;
    37         int len=vec[u][i].second;
    38         if(v==fa)   continue;
    39         dfs(v,u,len);
    40     }
    41     for(int i=0;i<query[u].size();i++)
    42     {
    43         int v=query[u][i].first;
    44         int id=query[u][i].second;
    45         if(vis[v]==1)
    46         {
    47             ans[id]=Find(v).len+Find(u).len;
    48         }
    49     }
    50     pre[u].f=fa;
    51     pre[u].len+=length;
    52 }
    53 int main()
    54 {
    55     int T;
    56     cin>>T;
    57     while(T--)
    58     {
    59         int n,q,x,y,k;
    60         cin>>n>>q;
    61         for(int i=0;i<n-1;i++)
    62         {
    63             cin>>x>>y>>k;
    64             vec[x].push_back({y,k});
    65             vec[y].push_back({x,k});
    66             vis[y]=1;
    67         }
    68         for(int i=0;i<q;i++)
    69         {
    70             cin>>x>>y;
    71             query[x].push_back({y,i});
    72             query[y].push_back({x,i});
    73         }
    74         for(int i=1;i<=n;i++)
    75         {
    76             if(vis[i]==0)
    77             {
    78                 dfs(i,i,0);
    79                 break;
    80             }
    81         }
    82         for(int i=0;i<q;i++)
    83             cout<<ans[i]<<endl;
    84         for(int i=0;i<=n;i++)
    85         {
    86             vec[i].clear();
    87             query[i].clear();
    88         }
    89         memset(pre,0,sizeof(pre));
    90         memset(vis,0,sizeof(vis));
    91     }
    92     return 0;
    93 }
    
    
    
    
    

     

    
    
    
    
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  • 原文地址:https://www.cnblogs.com/ISGuXing/p/7220656.html
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