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  • 1077. Travelling Tours

    1077. Travelling Tours

    Time limit: 1.0 second Memory limit: 64 MB
    There are N cities numbered from 1 to N (1 ≤ N ≤ 200) and M two-way roads connect them. There are at most one road between two cities. In summer holiday, members of DSAP Group want to make some traveling tours. Each tour is a route passes K different cities (K > 2) T1, T2, …, TK and return to T1. Your task is to help them make Ttours such that:
    1. Each of these T tours has at least a road that does not belong to (T−1) other tours.
    2. T is maximum.

    Input

    The first line of input contains N and M separated with white spaces. Then follow by M lines, each has two number H and T which means there is a road connect city Hand city T.

    Output

    You must output an integer number T — the maximum number of tours. If T > 0, then T lines followed, each describe a tour. The first number of each line is K — the amount of different cities in the tour, then K numbers which represent K cities in the tour.
    If there are more than one solution, you can output any of them.

    Sample

    inputoutput
    5 7
    1 2
    1 3
    1 4
    2 4
    2 3
    3 4
    5 4
    
    3
    3 1 2 4
    3 1 4 3
    4 1 2 3 4
    
    Problem Author: Nguyen Xuan My (Converted by Dinh Quang Hiep and Tran Nam Trung)  Problem Source: From the third contest at Department of Mathematics and Informatics - Natural Sciences College - National University of HaNoi.
    ***************************************************************************************
    并查集;
    每次求出一个环后,标记结点,直到找出所有的环
    ***************************************************************************************
     1 #include<iostream>
     2 #include<string>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<algorithm>
     6 #include<queue>
     7 #include<vector>
     8 using namespace std;
     9 int head[1001],fa[1001];
    10 int n,m,i,j,k,cnt;
    11 int link[1001][1001];//联系数组
    12 int  vis[1001];//标记结点所用
    13 vector<int>ans[180000];//存储结果
    14 void make()
    15 {
    16     for(int it=0;it<=n;it++)
    17      {
    18          fa[it]=it;
    19          link[it][0]=0;
    20          ans[it].clear();
    21      }
    22 }
    23 int find(int x)//并查集
    24  {
    25      if(fa[x]!=x)
    26         fa[x]=find(fa[x]);
    27      return fa[x];
    28  }
    29  void  work(int x,int y)
    30   {
    31       memset(vis,0,sizeof(vis));
    32       memset(head,-1,sizeof(head));//前驱存储
    33       head[x]=-1;
    34       vis[x]=1;
    35       queue<int>Q;
    36       Q.push(x);
    37       while(!Q.empty())
    38        {
    39            int h=Q.front();
    40            Q.pop();
    41            if(h==y)
    42             {
    43                 ans[cnt].push_back(y);
    44                 for(int it=head[y];it!=-1;it=head[it])
    45                  ans[cnt].push_back(it);
    46                  cnt++;
    47                  return;
    48             }
    49             for(int it=1;it<=link[h][0];it++)
    50              {
    51                  int gs=link[h][it];//把没访问的节点放入队列
    52                  if(!vis[gs])
    53                   {
    54                       Q.push(gs);
    55                       vis[gs]=1;
    56                       head[gs]=h;
    57                   }
    58              }
    59 
    60        }
    61 
    62 
    63   }
    64   int main()
    65   {
    66       cin>>n>>m;
    67       int x,y;
    68        cnt=0;
    69        make();
    70       for(i=1;i<=m;i++)
    71        {
    72            cin>>x>>y;
    73            int fs=find(x);
    74            int ds=find(y);
    75            if(fs==ds)
    76             {
    77                 work(x,y);
    78             }
    79            else
    80            {
    81                link[x][++link[x][0]]=y;
    82                link[y][++link[y][0]]=x;
    83                fa[fs]=ds;
    84            }
    85 
    86         }
    87     cout<<cnt<<endl;
    88     for(i=0;i<cnt;i++)
    89      {
    90          int hs=ans[i].size();
    91          cout<<hs<<' ';
    92          for(j=0;j<hs;j++)
    93             cout<<ans[i][j]<<' ';
    94          cout<<endl;
    95       }
    96     return 0;
    97 
    98   }
    View Code
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  • 原文地址:https://www.cnblogs.com/sdau--codeants/p/3283676.html
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