• How far away ?


    How far away ?

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

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 25399    Accepted Submission(s): 10108


    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
     

    在JV dalao的指导下,终于学会了tarjan。。。

    推荐一下他写的关于tarjan的文章,很详细:http://www.cnblogs.com/JVxie/p/4854719.html

    模板题

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <string>
     7 #include<vector>
     8 #include <queue>
     9 const int INF=0x3f3f3f3f;
    10 #define maxn 40005
    11 using namespace std;
    12 
    13 int n,m;
    14 
    15 vector<pair<int,int> >mp[maxn],book[maxn];
    16 int fa[maxn];
    17 int vis[maxn],dis[maxn];
    18 int ans[maxn];
    19 
    20 int Find(int x){
    21     int r=x,y;
    22     while(x!=fa[x]){
    23         x=fa[x];
    24     }
    25     while(r!=x){
    26         y=fa[r];
    27         fa[r]=x;
    28         r=y;
    29     }
    30     return x;
    31 }
    32 
    33 int join(int x,int y){
    34     int xx=Find(x);
    35     int yy=Find(y);
    36     if(xx!=yy){
    37         fa[yy]=xx;
    38         return true;
    39     }
    40     return false;
    41 }
    42 
    43 void tarjan(int x,int pre,int dist){
    44     int to,len,tmp;
    45     for(int i=0;i<mp[x].size();i++){
    46         to=mp[x][i].first;
    47         len=mp[x][i].second;
    48         tmp=dist+len;
    49         if(to!=pre&&!vis[to]){
    50             dis[to]=tmp;
    51             tarjan(to,x,tmp);
    52             vis[to]=1;
    53             join(x,to);
    54         }  
    55     }
    56     int pos;
    57     for(int i=0;i<book[x].size();i++){
    58         to=book[x][i].first;
    59         pos=book[x][i].second;
    60         if(vis[to]&&ans[pos]==-1){
    61             tmp=Find(to);
    62             ans[pos]=dis[to]+dis[x]-2*dis[tmp];
    63         }
    64     }
    65 }
    66 
    67 int main(){
    68     std::ios::sync_with_stdio(false);
    69     int T;
    70     cin>>T;
    71     while(T--){
    72         memset(vis,0,sizeof(vis));
    73         memset(ans,-1,sizeof(ans));
    74         memset(dis,0,sizeof(dis));
    75         for(int i=0;i<=n;i++){
    76             mp[i].clear();
    77             book[i].clear();
    78         }
    79         int x,y,z;
    80         cin>>n>>m;
    81         for(int i=0;i<=n;i++) fa[i]=i;
    82         for(int i=1;i<n;i++){
    83             cin>>x>>y>>z;
    84             mp[x].push_back(make_pair(y,z));
    85             mp[y].push_back(make_pair(x,z));
    86         }
    87         for(int i=1;i<=m;i++){
    88             cin>>x>>y;
    89             book[x].push_back(make_pair(y,i));
    90             book[y].push_back(make_pair(x,i));
    91         }
    92         tarjan(1,-1,0);
    93         for(int i=1;i<=m;i++){
    94             cout<<ans[i]<<endl;
    95         }
    96     }
    97     system("pause");
    98 }
    View Code
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  • 原文地址:https://www.cnblogs.com/Fighting-sh/p/10004427.html
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