PAT 1122 Hamiltonian Cycle

The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle".

In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2), the number of vertices, and M, the number of edges in an undirected graph. Then M lines follow, each describes an edge in the format Vertex1 Vertex2, where the vertices are numbered from 1 to N. The next line gives a positive integer K which is the number of queries, followed by K lines of queries, each in the format:

n V?1?? V?2?? ... V?n??

where n is the number of vertices in the list, and V?i??‘s are the vertices on a path.

Output Specification:

For each query, print in a line YES if the path does form a Hamiltonian cycle, or NO if not.

Sample Input:

6 10
6 2
3 4
1 5
2 5
3 1
4 1
1 6
6 3
1 2
4 5
6
7 5 1 4 3 6 2 5
6 5 1 4 3 6 2
9 6 2 1 6 3 4 5 2 6
4 1 2 5 1
7 6 1 3 4 5 2 6
7 6 1 2 5 4 3 1

Sample Output:

YES
NO
NO
NO
YES
NO

#include<iostream>
#include<vector>
#include<set>
using namespace std;
int main(){
  int n, m, i, j;
  scanf("%d%d", &n, &m);
  vector<set<int> > v(n+1);
  for(i=0; i<m; i++){
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].insert(b);
    v[b].insert(a);
  }
  int k;
  scanf("%d", &k);
  for(i=0; i<k; i++){
    int cnt, vertex;
    scanf("%d", &cnt);
    vector<int> qry(cnt), vis(n+1, false);
    for(j=0; j<cnt; j++) scanf("%d", &qry[j]);
    bool flag=false;
    if(cnt!=n+1 || qry[0]!=qry[cnt-1])flag=true;
    else{
      for(j=0; j<cnt-1; j++){
        if(!vis[qry[j]] && v[qry[j]].find(qry[j+1])!=v[qry[j]].end()) vis[qry[j]]=true;
        else{
          flag=true;
          break;
        }
      }
    }
     printf("%s\n", flag ? "NO" : "YES");
  }
  return 0;
}

原文地址：https://www.cnblogs.com/mr-stn/p/9571493.html