Question

我遇到了一个用于创建ascii迷宫的算法的问题。代码使用递归后向跟踪器，伪代码基本上是：

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

我遇到的问题是代码被卡在添加到堆栈然后弹出堆栈的循环中。如果有人需要为Windows机器更改它，代码当前在第19行使用一个Linux系统命令

#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

如果在isGood函数（grid [x] [y]而不是grid [y] [x]）中，您首先交换了x和y。这有时会导致问题。

递归回溯迷宫生成算法堆栈循环

1 个答案: