Step on the Square

step on the square

Problem Description

There is a grid matrix, and the matrix boundary is infinitely far away. Let us make the following assumptions:

A. Each step, you can only move one block from the current square and walk to an adjacent square;
C. Only go north, east and west in three directions;

Excuse me: If we allow n steps (n<=20) on the grid matrix, how many different schemes are there? 2 ways to go is considered a different scheme as long as there is a difference.

Ideas:
Recursive

Starting from (I,J), the number of steps to walk n is equal to the sum of the following three items:

Starting from (I+1,J), take the number of n-1 steps. Premise: (I+1,J) has not passed
Starting from (i,j+1), take the number of n-1 steps. Premise: (i,j+1) has not passed
Starting from (I,J-1), take the number of n-1 steps. Premise: (i,j-1) has not passed

 #include <iostream> #include <algorithm> #include <cstring> using

namespace Std;

int VISITED[30][50];
    int ways (int i, int j, int n) {if (n==0) return 1;
    VISITED[I][J] = 1;
    int num = 0;
    if (!visited[i+1][j]) num + = Ways (I+1, J, n-1);//Note Set can only be a North thing, so here is +1 if (!visited[i][j-1]) num + = ways (i, j-1, n-1);

    if (!visited[i][j+1]) num + = ways (i, j+1, n-1);
    VISITED[I][J] = 0;
return num;
    } int main () {int n;
    CIN >> N;
    memset (visited, 0, sizeof (visited));
    cout << Ways (0, N) <<endl;
return 0; }