我有一个解决迷宫的程序,以便找到从开始(S)到退出(E)的可能路线。这是我的迷宫:
1111S11110
0000010001
110100010d
t001111110
0100000001
0111111101
1111111101
00000D01T1
0111110001
0000E01110
可能的路线是:
Start S W W S S S E E E E E E S S S S W W N W W W W W W S S E E E E Exit
这是正确的,我在CodeBlocks上得到它,但是当我通过putty在dev上编译我的代码时,我得到了这个:
Start S S S S S W N N W S W N N N W N Exit
以下是我的全部代码:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
// Global Variables for use
int matrixSize,startX,startY,exitX,exitY;
char src[1500] = " ";
char Ndirection[50] = " N ";
char Sdirection[50] = " S ";
char Edirection[50] = " E ";
char Wdirection[50] = " W ";
// Function for finding the array length
int numOfLines(FILE *const mazeFile) {
int c, count;
count = 0;
for (;; ) {
c = fgetc(mazeFile);
if (c == EOF)
break;
if (c == '\n')
++count; // end of line => increment line counter
}
rewind(mazeFile);
return count+1;
}
int capLetter(char ch){
int result = 0;
if(ch >= 'A' && ch <= 'Z'){
result = 1;
}
return result;
}
int lowLetter(char ch){
int result = 0;
if(ch >= 'a' && ch <= 'z'){
result = 1;
}
return result;
}
int isSafe(char Mazearray[matrixSize][matrixSize],int x,int y){
if(x >= 0 && x < matrixSize && y >= 0 && y < matrixSize && Mazearray[x][y] != '1'){
return 1;
}
return 0;
}
void MazeSolution(char Mazearray[matrixSize][matrixSize],int x,int y,char pathArray[matrixSize][matrixSize],char wasHereArray[matrixSize][matrixSize]){
if(recursiveMaze(Mazearray,x,y,pathArray,wasHereArray) == 0){
printf("There does not exist a possible solution!!!");
}
else{
pathArray[startX][startY] = 'S';
}
}
int recursiveMaze(char Mazearray[matrixSize][matrixSize],int x,int y,char pathArray[matrixSize][matrixSize],char wasHereArray[matrixSize][matrixSize]){
if(x == startX && y == startY){
pathArray[x][y] = 'S';
}
if(x == exitX && y == exitY){
pathArray[x][y] = 'E';
return 1;
}
// check if the coordinate is safe to go(not 1)
if(isSafe(Mazearray,x,y) == 1 && wasHereArray[x][y] != '1'){
wasHereArray[x][y] = '1';
// Move North
if(recursiveMaze(Mazearray,x-1,y,pathArray,wasHereArray) == 1){
pathArray[x][y] = 'R';
strcat(src,Ndirection);
return 1;
}
// Move South
if(recursiveMaze(Mazearray,x+1,y,pathArray,wasHereArray) == 1){
pathArray[x][y] = 'R';
strcat(src,Sdirection);
return 1;
}
// Move East
if(recursiveMaze(Mazearray,x,y+1,pathArray,wasHereArray) == 1){
pathArray[x][y] = 'R';
strcat(src,Edirection);
return 1;
}
// Move West
if(recursiveMaze(Mazearray,x,y-1,pathArray,wasHereArray) == 1){
pathArray[x][y] = 'R';
strcat(src,Wdirection);
return 1;
}
}
return 0;
}
// Main Function
int main( int argc, char **argv )
{
// Opening the Matrix File
FILE *mazeFile;
mazeFile = fopen(argv[1], "r" );
if( mazeFile == NULL )
return 1;
matrixSize = numOfLines( mazeFile );
// Reading text file into 2D array
int i,j;
char mazeArray [matrixSize][matrixSize];
for (i = 0; i<matrixSize; i++) {
for (j = 0; j<matrixSize; j++) {
fscanf(mazeFile, "%c", &mazeArray[i][j]);
}
char eol;
fscanf(mazeFile, "%c", &eol);
}
// Variables
//Creating path array
char pathArray[matrixSize][matrixSize];
for (i = 0; i < matrixSize; i++){
for (j = 0; j < matrixSize; j++){
pathArray[i][j] = '0';
}
}
// CheckPoint array
char wasHereArray[matrixSize][matrixSize];
for (i = 0; i < matrixSize; i++){
for (j = 0; j < matrixSize; j++){
wasHereArray[i][j] = '0';
}
}
// Finding start and exit indexes
for (i = 0; i<matrixSize; i++) {
for (j = 0; j<matrixSize; j++) {
if(mazeArray[i][j] == 'S'){
startX = i;
startY = j;
}
if(mazeArray[i][j] == 'E'){
exitX = i;
exitY = j;
}
}
}
MazeSolution(mazeArray,startX,startY,pathArray,wasHereArray);
char *data = src;
int length=strlen(data);
char bytes[length];
int n=0;
while(n<=length)
{
bytes[n] = data[length-n-1];
n++;
}
FILE *f = fopen("path.txt", "w");
fprintf(f, "Start %s Exit",bytes);
fclose(mazeFile);
fclose(f);
return 0;
}
我不知道出了什么问题,从哪里开始?
答案 0 :(得分:1)
DOS行结尾是CR-LF(“\ r \ n”),* nix行结尾只是LF(“\ n”)。更改这些行main
:
char eol;
fscanf(mazeFile, "%c", &eol);
为:
int c = fgetc(mazefile); // Slurp a '\r' carriage return or '\n' linefeed character.
if ('\r' == c) {
c = fgetc(mazefile); // slurp the '\n' linefeed character.
}