给定一个表示N x N方阵的整数N ，任务是按照以下条件打印从方阵左上角到右下角的移动方式：
- 如果指针的当前位置在方阵的边缘，则下一个移动可以是垂直移动或水平移动，可以执行任意数量的步长(如棋盘上的车子)。
- 如果指针的当前位置在方矩阵的对角线上，则下一个移动也应位于对角线上。 (就像主教在棋盘上一样)。
- 如果指针的当前位置在上述两个位置之外的任何其他位置，则下一步可能会以任何方式进行，例如骑士在棋盘上移动，但是新位置的行和列>旧位置的行和列位置。
例子：
Input: N = 2
Output: 3
Explanation:
The three possible ways are:
{0-0} – {0-1} – {1-1}
{0-0} – {1-0} – {1-1}
{0-0} – {1-1}
Input: 3
Output: 18
Explanation:
The possible ways are:
{0-0} – {2-1} – {2-2}
{0-0} – {1-2} – {2-2}
{0-0} – {0-1} – {2-2}
{0-0} – {0-1} – {0-2} – {1-2} – {2-2}
{0-0} – {0-1} – {0-2} – {2-2}
{0-0} – {0-1} – {1-1} – {2-2}
{0-0} – {0-1} – {2-1} – {2-2}
{0-0} – {0-2} – {1-2} – {2-2}
{0-0} – {0-2} – {2-2}
{0-0} – {1-0} – {2-2}
{0-0} – {1-0} – {1-1} – {2-2}
{0-0} – {1-0} – {1-2} – {2-2}
{0-0} – {1-0} – {2-0} – {2-1} – {2-2}
{0-0} – {1-0} – {2-0} – {2-2}
{0-0} – {2-0} – {2-1} – {2-2}
{0-0} – {2-0} – {2-2}
{0-0} – {1-1} – {2-2}
{0-0} – {2-2}
方法：想法是在每种可能的情况下递归检查其是否到达平方矩阵的末尾。为此，定义了一个递归函数，该函数将当前位置和最后一个位置作为输入参数。以下是递归函数的条件：
- 基本情况：函数的基本情况是检查当前指针是否已到达方阵中的右下角位置。如果达到，则增加记下总数的计数器的计数。如果无法到达最后一个位置，则应停止该函数，并且不应在下一次迭代中调用该函数。这是通过以下方式实现的：
if (cr == er && cc == ec) {
count++;
return;
}
if (cr > er || cc > ec) {
return;
}
- 递归的情况：考虑到三种类型的遍历是可能的。因此，递归的情况是通过检查当前位置是否递归调用该函数。如果当前位置在方矩阵的边缘，则指针只能水平或垂直移动。对于随后的每个垂直遍历，还可以选择水平和垂直遍历的两种选择。因此，为了跟踪方法的总数，在循环中以递归方式调用了这两种方法。
for (int i = 1; i <= er; i++) {
if (cc == 0 || cr == 0
|| cr == er || cc == ec) {
chessboard1(cr, cc + i, er, ec);
}
}
for (int i = 1; i <= er; i++) {
if (cc == 0 || cr == 0
|| cr == er || cc == ec) {
chessboard1(cr + i, cc, er, ec);
}
}
- 除此之外，如果当前指针在对角线上，则指针可以沿对角线移动。然而，对于每个随后的对角线遍历，另一组后续的对角线是可能的。因此，为了跟踪方法的总数，使用了for循环来递归调用该函数。
for (int i = 1; i <= er; i++) {
if (cr == cc || cr + cc == er) {
chessboard1(cr + i, cc + i,
er, ec);
}
}
- 如果以上情况都不满足，则可以这样移动下一个位置：
- 新行是>旧行。
- 新列>旧列。
- 执行完上述函数，存储在count变量中的值就是遍历平方矩阵的方式总数。
下面是上述方法的实现：
C++
// C++ program to count the number
// of ways to traverse the given
// N x N Square Matrix recursively
#include
using namespace std;
// Variable to store the
// total number of ways to
// traverse the Square Matrix
// Function to recursively
// count the total number
// of ways to traverse the
// given N x N Square Matrix
void chessboard1(int cr, int cc,
int er, int ec,
int& count)
{
// If the last index has been
// reached, then the count is
// incremented and returned
if (cr == er && cc == ec)
{
count++;
return;
}
// If the last index cannot
// be reached
if (cr > er || cc > ec)
{
return;
}
// If the current position is
// neither on the edges nor
// on the diagonal, then the
// pointer moves like a knight
chessboard1(cr + 2, cc + 1,
er, ec,count);
chessboard1(cr + 1, cc + 2,
er, ec,count);
// If the pointer is on the
// edges of the Square Matrix
// next move can be horizontal
// or vertical
// This for loop is used to include
// all the horizontal traversal cases
// recursively
for (int i = 1; i <= er; i++)
{
if (cc == 0 || cr == 0||
cr == er || cc == ec)
{
chessboard1(cr, cc + i,
er, ec,count);
}
}
// This for loop is used to include
// all the vertical traversal cases
// recursively
for (int i = 1; i <= er; i++)
{
if (cc == 0 || cr == 0||
cr == er || cc == ec)
{
chessboard1(cr + i, cc,
er, ec,count);
}
}
// If the pointer is on the
// diagonal of the Square Matrix
// next move is also diagonal.
// For loop is used to include
// all the diagonal traversal cases.
for (int i = 1; i <= er; i++)
{
if (cr == cc || cr + cc == er)
{
chessboard1(cr + i, cc + i,
er, ec, count);
}
}
}
// Driver code
int main()
{
int N = 3;
int count=0;
chessboard1(0, 0, N - 1, N - 1,count);
cout< Java
// Java program to count the number
// of ways to traverse the given
// N x N Square Matrix recursively
public class GFG {
// Variable to store the total number
// of ways to traverse the Square Matrix
static int count = 0;
// Function to recursively
// count the total number
// of ways to traverse the
// given N x N Square Matrix
public static void chessboard1(
int cr, int cc,
int er, int ec)
{
// If the last index has been reached, then
// the count is incremented and returned
if (cr == er && cc == ec) {
count++;
return;
}
// If the last index cannot be reached
if (cr > er || cc > ec) {
return;
}
// If the current position is neither
// on the edges nor on the diagonal,
// then the pointer moves
// like a knight
chessboard1(cr + 2, cc + 1, er, ec);
chessboard1(cr + 1, cc + 2, er, ec);
// If the pointer is on the
// edges of the Square Matrix
// next move can be horizontal or vertical
// This for loop is used to include all the
// horizontal traversal cases recursively
for (int i = 1; i <= er; i++) {
if (cc == 0 || cr == 0
|| cr == er || cc == ec) {
chessboard1(cr, cc + i, er, ec);
}
}
// This for loop is used to include all the
// vertical traversal cases recursively
for (int i = 1; i <= er; i++) {
if (cc == 0 || cr == 0
|| cr == er || cc == ec) {
chessboard1(cr + i, cc, er, ec);
}
}
// If the pointer is on the
// diagonal of the Square Matrix
// next move is also diagonal.
// For loop is used to include
// all the diagonal traversal cases.
for (int i = 1; i <= er; i++) {
if (cr == cc
|| cr + cc == er) {
chessboard1(cr + i,
cc + i,
er, ec);
}
}
}
// Driver code
public static void main(String[] args)
{
int N = 3;
chessboard1(0, 0, N - 1, N - 1);
System.out.println(count);
}
}C#
// C# program to count the number
// of ways to traverse the given
// N x N Square Matrix recursively
using System;
class GFG{
// Variable to store the total number
// of ways to traverse the Square Matrix
static int count = 0;
// Function to recursively
// count the total number
// of ways to traverse the
// given N x N Square Matrix
public static void chessboard1(int cr, int cc,
int er, int ec)
{
// If the last index has been reached, then
// the count is incremented and returned
if (cr == er && cc == ec)
{
count++;
return;
}
// If the last index cannot be reached
if (cr > er || cc > ec)
{
return;
}
// If the current position is neither
// on the edges nor on the diagonal,
// then the pointer moves
// like a knight
chessboard1(cr + 2, cc + 1, er, ec);
chessboard1(cr + 1, cc + 2, er, ec);
// If the pointer is on the edges
// of the Square Matrix next move
// can be horizontal or vertical
// This for loop is used to include all the
// horizontal traversal cases recursively
for(int i = 1; i <= er; i++)
{
if (cc == 0 || cr == 0 ||
cr == er || cc == ec)
{
chessboard1(cr, cc + i, er, ec);
}
}
// This for loop is used to include all the
// vertical traversal cases recursively
for(int i = 1; i <= er; i++)
{
if (cc == 0 || cr == 0 ||
cr == er || cc == ec)
{
chessboard1(cr + i, cc, er, ec);
}
}
// If the pointer is on the
// diagonal of the Square Matrix
// next move is also diagonal.
// For loop is used to include
// all the diagonal traversal cases.
for(int i = 1; i <= er; i++)
{
if (cr == cc || cr + cc == er)
{
chessboard1(cr + i, cc + i,
er, ec);
}
}
}
// Driver code
public static void Main(string[] args)
{
int N = 3;
chessboard1(0, 0, N - 1, N - 1);
Console.Write(count);
}
}
// This code is contributed by rutvik_56输出 ：
18