迷宮求解

迷宮求解

其實問題很容易理解的，就是題目給出一個迷宮，入口位置和出口位置，求解該迷宮是否能夠走通。利用順序棧，逐個位置進行探索。

CODE：

#include <stdio.h>
#include <stdlib.h>
#define MAXSIZE 100
#define OK 1
#define FAIL 0
#define STACK_INIT_SIZE 10000

typedef int Status;

typedef struct{
    int x;
    int y;
}PosType;

typedef struct
{
    int ord;
    PosType seat;
    int di;
}SElemType;

typedef struct
{
    SElemType *base;
    SElemType *top;
    int stacksize;
}SqStack;



typedef struct{
    PosType seat;
    char pass;
    int visit;
}MazeType;

void InitStack(SqStack *S)
{
    S->base = (SElemType *)malloc(STACK_INIT_SIZE*sizeof(SElemType));
    S->top = S->base;
    S->stacksize = STACK_INIT_SIZE;
}

void Push(SqStack *S, SElemType e)
{
    *(S->top++) = e;
}

void Pop(SqStack *S, SElemType *e)
{
    *e = *(--S->top);
}

int EmptyStack(SqStack S)
{
    if(S.top == S.base)
        return 1;
    else
        return 0;
}

void NextPos(PosType *seat, int di)
{
    if(di == 1)
        seat->x = seat->x + 1;
    if(di == 2)
        seat->y = seat->y - 1;
    if(di == 3)
        seat->x = seat->x - 1;
    if(di ==4)
        seat->y = seat->y + 1;
}

SqStack S;

Status MazePath(MazeType maze[][MAXSIZE], PosType start, PosType end) //maze[][n], the first [] indicates it is a pointer
{

    SElemType e;
    InitStack(&S);
    PosType curpos;
    curpos = start;
    int curstep = 1;

    do
    {
        if((*(*(maze+curpos.x)+curpos.y)).pass == 'p' && (*(*(maze+curpos.x)+curpos.y)).visit == 0)
        {
            (*(*(maze+curpos.x)+curpos.y)).visit = 1;
            e.ord = curstep;
            e.seat = curpos;
            e.di = 1;
            Push(&S,e);
            if (curpos.x == end.x && curpos.y == end.y)
                return OK;
            NextPos(&curpos, 1);
            curstep++;
        }
        else
        {
            if(!EmptyStack(S))
            {
                Pop(&S,&e);
                while(e.di==4 && !EmptyStack(S))
                {
                    (*(*(maze+e.seat.x)+e.seat.y)).pass = 'b';
                    Pop(&S,&e);
                }
                if(e.di<4)
                {
                    e.di++;
                    Push(&S, e);
                    NextPos(&e.seat, e.di);
                    curpos = e.seat;
                }
            }
        }
    }while(!EmptyStack(S));
    return FAIL;
}



int main()
{
    int m, n, i, j;
    char c;
    PosType start, end;
    MazeType maze[MAXSIZE][MAXSIZE];
    printf("Please enter the number of lines of the maze: ");
    scanf("%d", &m);
    printf("Please enter the number of columns of the maze: ");
    scanf("%d", &n);
    printf("Please enter the maze pattern:\n"); // input format:bbbbbppbbppbbbbb b represents blocked, p represents pass
    //getchar();
    fflush(stdin);// refresh the buffer
    for(i=0; i<m; i++){
        for(j=0; j<n; j++){
            scanf("%c", &maze[i][j].pass);
            maze[i][j].seat.x = i;
            maze[i][j].seat.y = j;
            maze[i][j].visit = 0;
        }
        fflush(stdin);
    }
    printf("Please enter the position of the start point:");
    scanf("%d %d", &(start.x), &(start.y));
    printf("Please enter the position of the end point:");
    scanf("%d %d", &(end.x), &(end.y));
    SElemType e;
    if(MazePath(maze, start, end)){
        printf("There is a path O(∩_∩)O~\n");
        while(!EmptyStack(S)){
            Pop(&S, &e);
            printf("(%d, %d)->", e.seat.x, e.seat.y);
        }

    }
    else
        printf("This is a maze you will never get out!");

return 0;
}

• 有一段時間在調試的時候輸入總是出現問題，查了一下是scanf()其實是會接收回車在緩衝區的（以前貌似沒碰到過這種情況啊），所以加上fflush(stdin)刷新一下緩衝區就好了。
• 最後在輸出路徑的時候其實是從終點往起點輸出的，順序是反的。
• 形參用&S的引用形式編譯器會報錯，但是很多書上是這樣寫的，於是改成了指針作爲函數的參數。

這個迷宮做了還蠻長時間的，因爲彎彎繞繞的老是搞不清楚，看了很多參考資料，最後一步一步調試，在紙上慢慢抄慢慢調。沉住氣，堅持，你做的遠遠不夠。