遞歸回溯迷宮生成算法堆棧循環

Question

我遇到了一個我爲創建ascii迷宮而編寫的算法的問題。該代碼使用遞歸回到跟蹤器，以及僞代碼本質是：

1. Make the initial cell the current cell and mark it as visited 
2. While there are unvisited cells 
    1. If the current cell has any neighbours which have not been visited 
     1. Choose randomly one of the unvisited neighbours 
     2. Push the chosen cell to the stack 
     3. Remove the wall between the current cell and the chosen cell 
     4. Make the chosen cell the current cell and mark it as visited 
    2. Else 
     1. Pop a cell from the stack 
     2.Make it the current cell 

，我有是代碼被卡在增加了棧，然後出棧的循環問題。該代碼目前第19行使用一個系統命令的Linux，如果任何人需要改變它的Windows機器

#include<iostream> 
#include<cstdlib> 
#include<stack> 
#include<ctime> 

#define NORTH 0 
#define SOUTH 1 
#define EAST 2 
#define WEST 3 
#define SIZEX 20 
#define SIZEY 20 

using namespace std; 

int nGood = 0; 
int locX = 1, locY = 1; 

void printGrid(char grid[SIZEY][SIZEX]){ 
system("clear"); 
for (int i = 0; i < SIZEY; i++){ 
    for(int j = 0; j < SIZEX; j++){ 
     cout << grid[i][j]; 
    } 
    cout << endl; 
} 
} 

int moveEW(int direction, int x){ 
    if (direction == EAST) 
      return x + 1; 
    else if (direction == WEST) 
      return x - 1; 
    else 
      return x; 
} 

int moveNS(int direction, int y){ 
    if (direction == NORTH) 
      return y - 1; 
    else if (direction == SOUTH) 
      return y + 1; 
    else 
      return y; 
} 

bool isGood(int x, int y, int direction, char grid[SIZEY][SIZEX]){ 
x = moveEW(direction,x); 
y = moveNS(direction,y); 

if (grid[x][y] == '.' || x >= (SIZEX - 1) || x <= 0 || y <= 0 || y >= (SIZEY - 1)){ 
    return false; 
} 

// check cardinal directions 
if (direction == NORTH){ 
    if (grid[y][x-1] != '.' && grid[y-1][x] != '.' && grid[y][x+1] != '.' && grid[y-1][x-1] != '.' && grid[y-1][x+1] != '.'){ 
     return true; 
    } 
} 
if (direction == SOUTH){ 
      if (grid[y][x-1] != '.' && grid[y+1][x] != '.' && grid[y][x+1] != '.' && grid[y+1][x-1] != '.' && grid[y+1][x+1] != '.'){ 
        return true; 
      } 
    } 
if (direction == EAST){ 
      if (grid[y][x+1] != '.' && grid[y-1][x] != '.' && grid[y+1][x] != '.' && grid[y-1][x+1] != '.' && grid[y+1][x+1] != '.'){ 
        return true; 
      } 
    } 
if (direction == WEST){ 
      if (grid[y][x-1] != '.' && grid[y-1][x] != '.' && grid[y+1][x] != '.' && grid[y-1][x-1] != '.' && grid[y+1][x-1] != '.'){ 
        return true; 
      } 
    } 
return false; 
} 

main(){ 
char grid[SIZEY][SIZEX]; 

// init grid 
for (int i = 0; i < SIZEY; i++){ 
      for(int j = 0; j < SIZEX; j++){ 
        grid[i][j] = '#'; 
      } 
    } 

//init rand 
srand(time(0)); 

//init stacks for xy coords 
stack<int> xValues; 
stack<int> yValues; 

nGood = 0; 
int direction = 0; 

do{ 
    //find n good moves 
    for (int i = 0; i < 4; i++){ 
     if (isGood(locX,locY,i,grid)) 
      nGood++; 
    } 

    // if only 1 good move, move there 
    if (nGood == 1){ 
     if (isGood(locX,locY,NORTH,grid)) 
      locY = moveNS(NORTH,locY); 
     else if (isGood(locX,locY,SOUTH,grid)) 
      locY = moveNS(SOUTH,locY); 
     else if (isGood(locX,locY,EAST,grid)) 
          locX = moveEW(EAST,locX); 
        else if (isGood(locX,locY,WEST,grid)) 
          locX = moveEW(WEST,locX); 
    } 

    // if no good moves, move back in stack 
    else if (nGood == 0){ 
     locX = xValues.top(); 
     locY = yValues.top(); 
     xValues.pop(); 
     yValues.pop(); 
    } 

    //if more than 1 good move, push stack 
    else if (nGood > 1){ 
     xValues.push(locX); 
     yValues.push(locY); 

     //direction to move randomly chosen 
     do{ 
      direction = rand() % 4; 
     }while (!isGood(locX,locY,direction,grid)); 

     locX = moveEW(direction,locX); 
     locY = moveNS(direction,locY); 
    } 

    // set grid 
    grid[locY][locX] = '.'; 
    //output grid to show creation 
    printGrid(grid); 
    //reset nGood value 
      nGood = 0; 

}while(!xValues.empty()); 

//final maze output 
printGrid(grid); 
return 0; 
} 

Answer 1

您在isGood功能交換在第一X和Y如果（網格[X] [Y]代替格[Y] [X]）。這至少有時會導致問題。