/*
* 回溯法的典型问题
*/

/*
* 题目：设计一个函数，用来判断在一个矩形中是否存在一条包含
* 某字符串所有字符的路径。路径可以从矩阵中任意一格开始，
* 每一步可以在矩阵中向左右上下移动一格。一个格子不能经过
* 多次。
*/

#include<cstring>

bool hasPathCore(const char* matrix, int rows, int cols, int row, int col, const char* str, int& pathLength, bool* visited)
{
	if (str[pathLength] == '\0')
	{
		return true;
	}

	bool hasPath = false;
	if (row >= 0 && row < rows && col >= 0 && col < cols && matrix[row * cols + col] == str[pathLength] && !visited[row * cols + col])
	{
		++pathLength;
		visited[row * cols + col] = true;

		hasPath = hasPathCore(matrix, rows, cols, row, col-1, str, pathLength, visited)
			|| hasPathCore(matrix, rows, cols, row, col + 1, str, pathLength, visited)
			|| hasPathCore(matrix, rows, cols, row - 1, col, str, pathLength, visited)
			|| hasPathCore(matrix, rows, cols, row + 1, col, str, pathLength, visited);

		if (!hasPath)
		{
			--pathLength;
			visited[row * cols + col] = false;
		}
	}

	return hasPath;
}

bool hasPath(char* matrix, int rows, int cols, char* str)
{
	if (matrix == nullptr || rows < 1 || cols < 1 || str == nullptr)
	{
		return false;
	}

	bool* visited = new bool[rows * cols];
	memset(visited, 0, rows*cols);
	int pathLength = 0;
	for (int row = 0; row < rows; ++row)
	{
		for (int col = 0; col < cols; ++col)
		{
			if (hasPathCore(matrix, rows, cols, row, col, str, pathLength, visited))
			{
				delete[] visited;
				visited = nullptr;

				return true;
			}
		}
	}

	delete[] visited;
	visited = nullptr;

	return false;
}


/*
* 题目：在地上有一个m行n列的方格。一个机器人从坐标(0,0)的格子开始移动，
* 他每次可以向在左右上下移动一格，同一个格子不可以移动两次，也不能进入
* 行坐标和列坐标之和大于k的格子。
* 例如，当k为18时，机器人能够进入方格(35,37),因为3+5+3+7 = 18. 但它不能进入
* 方格(35, 38),因为3+5+3+8 = 19.请问该机器人能够到达多少个格子。
*/

/*
* 感觉上只需要遍历整个方格的下标，判断下标与k的关系就可以解答了。但是现在练习回溯法，
* 就用回溯法的思想解答。
*/

#include<cstring>

int getDigitSum(int number)
{
	int sum = 0;
	while (number > 0)
	{
		sum += number % 10;
		number /= 10;
	}

	return sum;
}

bool check(int threshold, int rows, int cols, int row, int col, bool* visited)
{
	if (row >= 0 && row<rows && col >=0 && col <cols && getDigitSum(row) + getDigitSum(col) <= threshold)
	{
		return true;
	}

	return false;
}

int movingCountCore(int threshold, int rows, int cols, int row, int col, bool* visited)
{
	int count = 0;
	if (check(threshold, rows, cols, row, col, visited))
	{
		visited[row * cols + col] = true;

		count = 1 + movingCountCore(threshold, rows, cols, row + 1, col, visited) +
			movingCountCore(threshold, rows, cols, row - 1, col, visited) +
			movingCountCore(threshold, rows, cols, row, col + 1, visited) +
			movingCountCore(threshold, rows, cols, row, col - 1, visited);
	}
	return count;
}

int movingCount(int threshold, int rows, int cols)
{
	if (threshold < 0 || rows <= 0 || cols <= 0)
	{
		return 0;
	}

	bool* visited = new bool[rows * cols];
	memset(visited, 0, rows * cols);

	int count = movingCountCore(threshold, rows, cols, 0, 0, visited);

	delete[] visited;
	visited = nullptr;

	return count;
}