马踏棋盘优化

之前做了个深搜遍历求解的方法,未优化,一个下午没有结果,而且代码尚不够精简

看了一下优化方法,又重做了一个,基本不需要时间就能得到一个解

一共一百来行,沾沾自喜一下

#include<stdio.h>
#include<windows.h>

#define MAX 8
#define NONE 0
#define START 1
#define INFINITE 88888

int chess[MAX][MAX];
int gowhere[][2] = { { 2, -1 }, { 2, 1 }, { -2, 1 }, { -2, -1 }, { 1, 2 }, { 1, -2 }, { -1, -2 }, { -1, 2 } };//用二维数组更简单
void initchess();
void traverchess(int i, int j, int num);
bool judge(int i, int j);
int judgeminpath(int i, int j);

int main()
{
	initchess();
	int i, j;
	scanf_s("%d%d", &i, &j);
	traverchess(i, j, START);
	system("pause");
}

void initchess()
{
	int i, j;
	for (i = 0; i < MAX; i++)
	{
		for (j = 0; j < MAX; j++)
		{
			chess[i][j] = NONE;
		}
	}
}

void traverchess(int i, int j,int num)
{
	if (num == MAX*MAX)
	{
		int x, y;
		chess[i][j] = MAX*MAX;
		for (x = 0; x < MAX; x++)
		{
			for (y = 0; y < MAX; y++)
			{
				printf("%4d   ", chess[x][y]);
			}
			printf("\n");
		}
		printf("\n");
		return;
	}
	int recordc,count,min=INFINITE;
	for (count = 0; count < 8; count++)//for循环为贪心算法核心,遍历坐标ij的周围八个点,并由judgeminpath函数返回八个点的中每个点的可走路数
	{
		if (!judge(i + gowhere[count][0], j + gowhere[count][1]))
		{
			continue;
		}
		int path = judgeminpath(i + gowhere[count][0], j + gowhere[count][1]);
		if (min > path)
		{
			min = path;
			recordc = count;
		}
	}
	if (min!=INFINITE)
	{
		chess[i][j] = num;
		traverchess(i + gowhere[recordc][0], j + gowhere[recordc][1], num + 1);
		chess[i][j] = NONE;
	}
}

bool judge(int i,int j)
{
	if (i >= MAX || i < 0 || j >= MAX || j < 0)
	{
		return false;
	}
	if (chess[i][j] == NONE)
	{
		return true;
	}
	else
	{
		return false;
	}
}

int judgeminpath(int i, int j)
{
	int c;
	int tag=0;
	for (c = 0; c < 8; c++)
	{
		if (judge(i + gowhere[c][0], j + gowhere[c][1]))
		{
			tag++;
		}
	}
	return tag;
}

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转载自blog.csdn.net/qq_42705770/article/details/83855664