The realization of C language by recursive + backtracking method of horse stepping checkerboard algorithm

R is the row of the Matrix, and C is the column of the matrix

Output the results to the results.txt in the current directory (need to be built in advance).

The results are given: 1. The existence of a path allows the horse to travel all squares as required;

2. Total number of solutions;

3. Time of execution of the program;

#include <stdio.h>#include<stdlib.h>#include<time.h>#defineR2#defineC 4intflag[r][c]={0};//a two-dimensional array that holds the horse jumping path    intarr[r][c]={0}; intx[8]={2,1,-1,-2,2,1,-1,-2}; inty[8]={1,2,2,1,-1,-2,-2,-1}; intn=1, Count; FILE*p; //initializing the file output stream    intinitout () {FP=fopen ("Results.txt","W");//Remember to close        if(fp==NULL) {printf ("File cannot open!" ); Exit (0); }        return 0; }        //Add a judgment function to determine if such a Hamiltonian path exists    voidjudgeexistence () {if(count==0) printf ("a chessboard of%d *%d does not exist that will allow the horse to traverse the path of each of the squares without repeating \ n", R,c); }    //output Horse Jump step stepout ()    voidStepOut () { for(intA=0; a<r;a++){                  for(intb=0; b<c;b++) {fprintf (FP),"%3d", Arr[a][b]); } fprintf (FP,"\ n"); } fprintf (FP,"\ n"); }        voidDFS (intIintj) {if(n==r*c) {count++; StepOut ();//write a function, output horse jump step stepout ()        }                Else           for(intk=0;k<8; k++)            {              if(i+x[k]>=0&&i+x[k]<r&&j+y[k]>=0&&j+y[k]<c&&flag[i+x[k]][j+y[k]]==0) {Flag[i+x[k]][j+y[k]]=1; n++; Arr[i+x[k]][j+y[k]]=n;//matrix assignment to record horse jumping stepsDFS (i+x[k],j+y[k]);//Loop + recursionFlag[i+x[k]][j+y[k]]=0; n--; }          }    }      voidMain () {clock_t start, finish; //How long does it take to calculate the program?        Longduration; Start=clock (); Count=0; intx, y; label_1:printf ("Please enter the horse's initial horizontal axis (x<=%d): x=\n", R); scanf ("%d",&X); if(x>R) {printf ("Please enter a number less than or equal to%d \ n", R); GotoLabel_1; } label_2:printf ("Please enter the horse initial ordinate (y<=%d): y=\n", c); scanf ("%d",&Y); if(y>B) {printf ("Please enter a number less than or equal to%d \ n", c); Gotolabel_2; } X=x-1; Y=y-1; Flag[x][y]=1; Arr[x][y]=1;        Initout ();        DFS (x, y);        Judgeexistence (); fprintf (FP,"The total number of solutions is:%d\n", Count); Finish=clock (); Duration=finish-start;//program execution time, per millisecondfprintf (FP,"the execution time of the program is:%10LD ms\n", duration);     Fclose (FP); }

What is wrong in the code welcome you to correct me.

The realization of C language by recursive + backtracking method of horse stepping checkerboard algorithm