哈密顿循环是哈密顿路径，因此从哈密顿路径的最后一个顶点到第一个顶点有一条边(在图中)。它在无向图中是一条路径，该路径恰好访问该图的每个顶点一次。
功能和目的：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
示例#include
#include
#include
#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
if (path[i] == n)
return false;
return true;
}
bool hamiltonianCycle(bool g[N][N], int path[], int pos) {
//如果所有顶点都包含在哈密顿循环中
if (pos == N) {
if (g[ path[pos-1] ][ path[0] ] == 1)
return true;
else
return false;
}
for (int n = 1; n
if (isSafe(n, g, path, pos)) //Check if this vertex can be added to Hamiltonian Cycle
{
path[pos] = n;
//重复构建其余路径
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
path[i] = -1;
//将顶点0作为路径中的第一个顶点。
If there is a Hamiltonian Cycle, then the path can be started from any point
//图是无向的循环周期
path[0] = 0;
if (hamiltonianCycle(g, path, 1) == false) {
cout<
return false;
}
displaytheSolution(path);
return true;
}
void displaytheSolution(int p[]) {
cout<
cout<
for (int i = 0; i
cout<
cout<
}
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;
}
输出结果存在周期： Following is one Hamiltonian Cycle
0 4 1 2 3 0
)
