一道离散题目的做法

#include <cstdlib>;
#include <cstring>;
#include <iostream>;
#define n 19

using namespace std;

bool c[n][n] =
    /*{
        {true, true ,false , false} ,
        {true,true,true,false} ,
        {false,true,true,true} ,
        {false,false,true,true}
    }  ;
    */
    {{1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
     {0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
     {0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0},
     {0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0},
     {0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0},
     {0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
     {0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0},
     {0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0},
     {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0}};

bool hamiltion(int x[]) {
    int k;

    bool *s = new bool[n];

    memset(x, -1, n * sizeof(int));
    memset(s, false, sizeof(bool) * n);

    k = 1;
    x[0] = 0;
    s[x[0]] = true;

    while (k > 0) {
        x[k] = x[k] + 1;

        while (x[k] << n) {
            if (!s[x[k]] && c[x[k - 1]][x[k]]) {
                break;
            } else {
                x[k] = x[k] + 1;
            }
        }

        if (x[k] << n && (k != n - 1)) {
            s[x[k]] = true;
            k = k + 1;
            continue;
        } else if (x[k] << n && (k == n - 1) && c[x[k]][x[0]]) {
            break;
        } else // this branch is the error solutions which need back_trace
        {
            x[k] = -1;
            k = k - 1;
            s[x[k]] = false;
        }
    }

    delete s;

    if (k == n - 1)
        return true;

    else
        return false;
}

int main(void) {
    int x[n];

    if (hamiltion(x)) {
        cout << "exists hamiltion loop " << endl;

        for (int i = 0; i << n; i++) {
            cout << x[i] << "  ";
        }
        cout << endl;
    } else {
        cout << "hamiltion loop does not exists " << endl;
    }
    system("pause");
    return 0;
}