一个生成数独的程序

标签：c++   程序员   面试   

以下是一个生成数独的程序，利用深度优先遍历的方式。当生成9x9的的数独时，我的个人电脑需要花费的时间太长，而3x3和6x6的可以正确生成。

//mySIZE是数独棋盘的边长，棋盘是mySIZE*mySIZE的大小
int mySIZE = 9;
void print(const vector<vector<int>> &num)
{
	for (int i = 0; i < mySIZE; i++)
	{
		for (int j = 0; j < mySIZE; j++)
		{
			cout << num[i][j] << "\t";
		}
		cout << endl;
	}
}

bool IsRightPlace(vector<vector<int>> &num, int row, int col)
{
	int n = num[row][col];
	for (int i = 0; i < mySIZE; i++)
	{
		if (i == row)
			continue;
		if (num[i][col] == n)
			return false;
	}
	for (int i = 0; i < mySIZE; i++)
	{
		if (i == col)
			continue;
		if (num[row][i] == n)
			return false;
	}
	int row_start = row / 3;
	row_start *= 3;
	int row_end = row_start + 2;
	int col_start = col / 3;
	col_start *= 3;
	int col_end = col_start + 2;
	int i = row_start, j = col_start;
	for (int k = 1; k <= 9; k++)
	{
		if (row != i || col != j)
		{
			if (num[i][j] == n)
				return false;
		}		
		if (j == col_end)
		{
			j = 0;
			i = i + 1;
		}
		else
		{
			j = j + 1;
		}
	}
	return true;
}

bool generate_core(vector<vector<int>> &num, int row, int col)
{
	
	vector<int> number;
	for (int i = 1; i <= 9; i++)
		number.emplace_back(i);
	while (!number.empty())
	{
		int randindex = rand() % number.size();
		int randnum = number[randindex];
		number.erase(number.begin() + randindex);
		num[row][col] = randnum;
		if (IsRightPlace(num, row, col) == false)
			continue;
		if (row == mySIZE - 1 && col == mySIZE-1)
		{
			return true;
		}
		int nextrow, nextcol;
		if (col == mySIZE-1)
		{
			nextrow=row + 1;
			nextcol = 0;
		}
		else
		{
			nextrow = row;
			nextcol = col + 1;
		}
		bool next = generate_core(num, nextrow, nextcol);
		if (next)
			return true;
	}
	if (number.empty())
	{
		num[row][col] = -5;
		return false;
	}		
}

void generate()
{
	vector<vector<int>> num(mySIZE, vector<int>(mySIZE, -1));
	generate_core(num, 0, 0);
	print(num);
}

一个生成数独的程序

标签：c++   程序员   面试   

原文地址：http://blog.csdn.net/bupt8846/article/details/43503447