PAT 1122 Hamiltonian Cycle（25 分）

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 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

思路：哈密顿回路：（1）构成环，首尾节点一样（2）覆盖所有点，并且除了第一个节点，其他节点只出现一次

程序：

#include <cstdio>
#include <vector>
#include <set>
using namespace std;
int edge[201][201];
int n,m;
bool isHamilton(vector<int> v)
{
  int size = v.size();
  set<int> temp;
  if(size != n + 1)
    return false;
  if(v[0] != v[size-1] && size > 1)
    return false;
  for(int i = 0; i < size - 1; i++)
  {
    if(!edge[v[i]][v[i+1]])
      return false;
    if(temp.count(v[i]) != 0)
      return false;
    temp.insert(v[i]);
  }
  return true;
}
int main()
{
  scanf("%d%d",&n,&m);
  for(int i = 0; i < m; i++)
  {
    int a,b;
    scanf("%d%d",&a,&b);
    edge[a][b] = edge[b][a] = 1;
  }
  int k;
  scanf("%d",&k);
  for(int i = 0; i < k; i++)
  {
    int nn;
    scanf("%d",&nn);
    vector<int> v;
    for(int i = 0; i < nn; i++)
    {
      int vex;
      scanf("%d",&vex);
      v.push_back(vex);
    }
    if(isHamilton(v))
      printf("YESn");
    else
      printf("NOn");
  }
  return 0;
}

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