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:
- 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.
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并查集；
每次求出一个环后，标记结点，直到找出所有的环
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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 }