与えられたグラフがハミルトン閉路を含まなければならないかどうかをチェックするC++プログラム

ハミルトン閉路は、ハミルトン経路の最後の頂点から最初の頂点までのエッジ（グラフ内）があるようなハミルトン経路です。無向グラフにあるのは、グラフの各頂点に1回だけアクセスするパスです。

機能と目的

Begin
   1. function isSafe() is used to check for whether it is
   adjacent to the previously added vertex and already not added.
   2. function hamiltonianCycle() solves the hamiltonian problem.
   3. function hamCycle() uses hamiltonianCycle() to solve the hamiltonian problem. It returns false if there is no
   Hamiltonian Cycle possible, otherwise return true and
   prints the path.
End

例

#include <iostream>
#include <cstdio>
#include <cstdlib>
#define N 5
using namespace std;
void displaytheSolution(int path[]);
bool isSafe(int n, bool g[N][N], int path[], int pos)
{
   if (g [path[pos-1]][n] == 0)
      return false;
   for (int i = 0; i < pos; i++)
      if (path[i] == n)
   return false;
   return true;
}
bool hamiltonianCycle(bool g[N][N], int path[], int pos)
{
   //If all vertices are included in Hamiltonian Cycle
   if (pos == N)
   {
      if (g[ path[pos-1] ][ path[0] ] == 1)
         return true;
      else
         return false;
   }
   for (int n = 1; n < N; n++)
   {
      if (isSafe(n, g, path, pos)) //Check if this vertex can be added to Hamiltonian Cycle
      {
         path[pos] = n;
         //recur to construct rest of the path
         if (hamiltonianCycle (g, path, pos+1) == true)
            return true;
         path[pos] = -1; //remove vertex if it doesn’t lead to the solution
      }
   }
   return false;
}
bool hamCycle(bool g[N][N])
{
   int *path = new int[N];
   for (int i = 0; i < N; i++)
      path[i] = -1;
   //put vertex 0 as the first vertex in the path. If there is a Hamiltonian Cycle, then the path can be started from any point
   //of the cycle as the graph is undirected
   path[0] = 0;
   if (hamiltonianCycle(g, path, 1) == false)
   {
      cout<<"\nCycle does not exist"<<endl;
      return false;
   }
   displaytheSolution(path);
   return true;
}
void displaytheSolution(int p[])
{
   cout<<"Cycle Exists:";
   cout<<" Following is one Hamiltonian Cycle \n"<<endl;
   for (int i = 0; i < N; i++)
      cout<<p[i]<<" ";
   cout<< p[0]<<endl;
}
int main()
{
   bool g[N][N] = {{0, 1, 0, 1, 1},
      {0, 0, 1, 1, 0},
      {0, 1, 0, 1, 1},
      {1, 1, 1, 0, 1},
      {0, 1, 1, 0, 0},
   };
   hamCycle(g);
   return 0;
}

出力

Cycle Exists: Following is one Hamiltonian Cycle
0 4 1 2 3 0