骑士周游

骑士周游问题求解

最新推荐文章于 2024-02-14 21:03:33 发布

转载最新推荐文章于 2024-02-14 21:03:33 发布 · 106 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/xiaochi/p/5035631.html

本文介绍了一个使用回溯法解决骑士周游问题的C++程序实现。通过定义一个N×N的棋盘，该程序从指定位置开始，寻找骑士访问棋盘上每个格子恰好一次的所有可能路径。

#include <iostream>
#include <iomanip>
using namespace std;

#define N 6

int a[N][N] = { 0 };

bool judge(int a[][N], int b, int c, int dir)
{
    if (dir == 0)
    {
        if (b - 2 >= 0 && c - 1 >= 0 && a[b - 2][c - 1] == 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }

    if (dir == 1)
    {
        if (b - 1 >= 0 && c - 2 >= 0 && a[b - 1][c - 2] == 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }

    if (dir == 2)
    {
        if (b - 2 >= 0 && c + 1 <= N-1 && a[b - 2][c + 1] == 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }

    if (dir == 3)
    {
        if (b - 1 >= 0 && c + 2 <= N-1 && a[b - 1][c + 2] == 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }

    if (dir == 4)
    {
        if (b + 2 <= N-1 && c - 1 >= 0 && a[b + 2][c - 1] == 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }

    if (dir == 5)
    {
        if (b + 1 <= N-1 && c - 2 >= 0 && a[b + 1][c - 2] == 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }

    if (dir == 6)
    {
        if (b + 2 <= N-1 && c + 1 <= N-1 && a[b + 2][c + 1] == 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }

    if (dir == 7)
    {
        //cout << "dir = 7\n";
        if (b + 1 <= N-1 && c + 2 <= N-1 && a[b + 1][c + 2] == 0)
        {
            return true;
        }
        else
        {
            return false;
        }
    }
}

void print()
{
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            cout << setw(5) << a[i][j] ;
        }
        cout << endl;
    }
}

int count1 = 1;
int count2 = 1;

void fun(int a[][N],int b,int c)
{
    if (count1 == N*N )
    {
        print();
        cout << "方案:" << count2++ << endl;
        cout << endl;
        system("pause");
    }
    else
    {
        for (int i = 0; i < 8; i++)
        {
            if (judge(a, b, c, i))
            {
                switch (i)
                {
                case 0:
                    count1++;
                    a[b - 2][c - 1] = count1;
                    fun(a, b - 2, c - 1);
                    a[b - 2][c - 1] = 0;
                    count1--;
                    break;

                case 1:
                    count1++;
                    a[b - 1][c - 2] = count1;
                    fun(a, b - 1, c - 2);
                    a[b - 1][c - 2] = 0;
                    count1--;
                    break;

                case 2:
                    count1++;
                    a[b - 2][c + 1] = count1;
                    fun(a, b - 2, c + 1);
                    a[b - 2][c + 1] = 0;
                    count1--;
                    break;

                case 3:
                    count1++;
                    a[b - 1][c + 2] = count1;
                    fun(a, b - 1, c + 2);
                    a[b - 1][c + 2] = 0;
                    count1--;
                    break;

                case 4:
                    count1++;
                    a[b + 2][c - 1] = count1;
                    fun(a, b + 2, c - 1);
                    a[b + 2][c - 1] = 0;
                    count1--;
                    break;

                case 5:
                    count1++;
                    a[b + 1][c - 2] = count1;
                    fun(a, b + 1, c - 2);
                    a[b + 1][c - 2] = 0;
                    count1--;
                    break;

                case 6:
                    count1++;
                    a[b + 2][c + 1] = count1;
                    fun(a, b + 2, c + 1);
                    a[b + 2][c + 1] = 0;
                    count1--;
                    break;

                case 7:
                    count1++;
                    a[b + 1][c + 2] = count1;
                    fun(a, b + 1, c + 2);
                    a[b + 1][c + 2] = 0;
                    count1--;
                    break;
                }
            }
        }
    }

}

void main()
{
    a[2][0] = 1;
    fun(a, 2, 0);

    system("pause");
}