PAT甲級1122

1122. Hamiltonian Cycle (25)

時間限制

300 ms

記憶體限制

65536 kB

代碼長度限制

16000 B

判題程式

Standard

作者

CHEN, Yue

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< N <= 200), 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 V1 V2 ... Vn

where n is the number of vertices in the list, and Vi'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<cstdio>
#include<set>
#include<algorithm>
using namespace std;
const int maxn = 210;
const int INF = 1000000000;
int a[maxn];
int G[maxn][maxn];
int main()
{
  fill(G[0], G[0] + maxn*maxn, INF);
  int N, M;
  scanf("%d %d", &N, &M);
  int v1, v2;
  for (int i = 0; i < M; i++)
  {
    scanf("%d %d", &v1, &v2);
    G[v1][v2] = 1;
    G[v2][v1] = 1;
  }
  int K;
  scanf("%d", &K);
  int n;
  set<int> s; bool flag = false;
  for (int i = 0; i < K; i++)
  {
    scanf("%d", &n);
    
    for (int j = 0; j < n; j++)
    {
      scanf("%d", &a[j]);
      s.insert(a[j]);
    }
    if (n != N + 1 || s.size() != N)
    {
      printf("NO\n");
    }
    else
    {
      int j = 0;
      if (n == N + 1 && s.size() == N)
      {
        for (j = 0; j < n - 1; j++)
        {
          if (G[a[j]][a[j + 1]] == 1 && a[0] == a[n - 1])
            continue;
          else
          {
            printf("NO\n");
            break;
          }
        }
        if (j == n - 1)
        {
          printf("YES\n");
        }
      }
    }
    s.clear();
  }
  return 0;
}