dfs--cell problem

Topic: 1009: Cell
Description
A rectangular array is made up of numbers 0 through 9, and numbers 1 to 9 represent cells.
The cell is defined as the same cell, along the cell number or the cell number.
The number of cells in the given rectangular array is obtained.
such as array
4 10
0234500067
1034560500
2045600671
0000000089
There are 4 cells.
Input
The first row contains 2 integers, n, and M, representing the number of rows and columns of the rectangular array, respectively.
The next n rows, each row m number, represent the cell's rectangular array.
Output
The number of output cells.
Sample Input
4 10
0234500067
1034560500
2045600671
0000000089
Sample Output
4 This problem can be solved with DFS and BFS, where DFS DFS code is used first:

Main.cpp//dfs--cell Issue///Created by Showlo on 2018/4/16. COPYRIGHT©2018 year Showlo.
All rights reserved.
#include <stdio.h> #include <algorithm> using namespace std;
#define MAX 1000 char Map[max][max];
int Vis[max][max];

int m,n;
    void Dfs (int x,int y) {int i,nx,ny;
    int dx[4]={1,0,-1,0};
    int dy[4]={0,1,0,-1};
        for (i=0; i<4; i++) {nx=x+dx[i];
       Ny=y+dy[i];
        printf ("%d%d\n", nx,ny); if (nx>=0&&nx<m&&ny>=0&&ny<n&&map[nx][ny]!= ' 0 ' &&vis[nx][ny]
            ==0) {//If added condition: &map[nx][ny]==map[x][y], you can calculate the number of cells under the same condition//printf ("%d%d\n", nx,ny);
            Vis[nx][ny]=1;
        DFS (NX,NY);
    else continue;
} return;
    int main () {int i,j,num;
        while (scanf ("%d%d", &m,&n)!=eof) {num=0;
        memset (Vis, 0, sizeof (VIS));
        For (i=0 i<m; i++) {scanf ("%s", Map[i]);
 }       For (i=0 i<m; i++) {for (j=0; j<n; J + +) {if (vis[i][j]==0&&map[i][j]!=
                    ' 0 ') {printf ("%d%d\n\n", i,j);
                    num++;
                    Vis[i][j]=1;
                DFS (I,J);
            else continue;
    } printf ("%d\n", num);
return 0; }

BFS Code: