1077. Travelling Tours

2023-10-14 07:40

文章标签 1077 travelling tours

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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) T ₁, T ₂, …, T_K and return to T ₁. Your task is to help them make Ttours such that:

Each of these T tours has at least a road that does not belong to (T−1) other tours.
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

input	output
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   }