Solution of 1122. Hamiltonian Cycle (25)

1122. Hamiltonian Cycle (25)

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

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结题思路 ：
题意要求在已知的无向图上，判断一个哈密尔顿回路是否成立。
即：无向图中的每个节点都只能经过一次，这也就意味着每条边最多只能经过一次。

程序步骤：
第一步、用一位数组保存无向图。
第二步、判断连续的走向对应的边是存在的，对应的节点是未访问的。

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具体程序(AC)如下：

#include <iostream>
#include <cstring>
using namespace std;
int main()
{
  bool map[40205];
  memset(map, sizeof(map), 0);
  int v, e;
  cin >> v >> e;
  int start, end;
  for(int i = 0; i < e; ++i)
  {
      cin>> start >> end;
      map[start * v + end] = map[end * v + start] = 1;
  }
  int n, jump, pre, cur;
  int i;
  int path[205];
  int node[205];
  cin>> n;
  while(n--)
  {
      memset(node, 0, sizeof(node));
      cin >> jump;
      for(i = 0; i< jump; ++i)
          cin >> path[i];
      if(jump != v + 1)
      {
          cout<<"NO"<<endl;
          continue;
      }
      if(path[0] != path[jump - 1])
      {
          cout<<"NO"<<endl;
          continue;
      }
      pre = path[0];
      for(i = 1; i < jump; ++i)
      {
          cur = path[i];
          if(map[cur*v+pre]||map[pre*v+cur])
          {
            if(node[cur])
                break;
            node[cur] = 1;
            pre = cur;  
          }
          else
              break;
      }
      if(i >= jump)
          cout<<"YES"<<endl;
      else
          cout<<"NO"<<endl;
  }
  return 0;
}