dfs求全排列相 只需要路径长度为n即可

int n, m;
bool st[N];
int path[N];
int dfs(int x,int state,int start)
{
    if(x == m)
    {
        for(int i = 0; i < n; i ++ )
        {
            cout << path[i] << " ";
        }
        cout << endl;
    }
    for(int i = 1; i <= n; i ++ )
    {
        if(!(state >> i & 1))
            {
                path[x] = i;
                dfs(x + 1, state | 1 << i, i + 1);
            }
    }
}
int main()
{
    cin >> n >> m;
    dfs(0,0,1);
}

n个数选m个全排列 修改路径长度即可

int n, m;
bool st[N];
int path[N];
int dfs(int x,int state,int start)
{
    if(x == m)
    {
        for(int i = 0; i < m; i ++ )
        {
            cout << path[i] << " ";
        }
        cout << endl;
    }
    for(int i = 1; i <= n; i ++ )
    {
        if(!(state >> i & 1))
            {
                path[x] = i;
                dfs(x + 1, state | 1 << i, i + 1);
            }
    }
}
int main()
{
    cin >> n >> m;
    dfs(0,0,1);
}

n个数选m个要求升序 只需要每次从start循环

int n, m;
bool st[26];
int path[26];
int dfs(int x,int state,int start)
{
    if(x == m)
    {
        for(int i = 0; i < m; i ++ )
        {
            cout << path[i] << " ";
        }
        cout << endl;
    }
    for(int i = start; i <= n; i ++ )
    {
        if(!(state >> i & 1))
            {
                path[x] = i;
                dfs(x + 1, state | 1 << i, i + 1);
            }
    }
}
int main()
{
    cin >> n >> m;
    dfs(0,0,1);
}