C++经典算法题-老鼠走迷官（一）

#include <stdio.h> 
#include <stdlib.h>

    int visit(int, int);

    int maze[7][7] = {{2, 2, 2, 2, 2, 2, 2},
        {2, 0, 0, 0, 0, 0, 2},
        {2, 0, 2, 0, 2, 0, 2},
        {2, 0, 0, 2, 0, 2, 2},
        {2, 2, 0, 2, 0, 2, 2},
        {2, 0, 0, 0, 0, 0, 2},
        {2, 2, 2, 2, 2, 2, 2}};

    int startI = 1, startJ = 1;		// 入口int endI = 5, endJ = 5;	//  出口int success = 0;

    int main(void) { int i, j;

        printf("显示迷宫：\n"); for(i = 0; i < 7; i++) {
            for(j = 0; j < 7; j++) if(maze[i][j] == 2)


                else

            printf("█");

            printf("	");

            printf("\n");
        }

        if(visit(startI, startJ) == 0)
            printf("\n没有找到出口！\n"); else {
            printf("\n显示路径：\n"); for(i = 0; i < 7; i++) {
                for(j = 0; j < 7; j++) { if(maze[i][j] == 2)
                    printf("█");
                else if(maze[i][j] == 1) printf("◇");

                else

                }


                printf("	");

                printf("\n");
            }
        }

        return 0;
    }

    int visit(int i, int j) { maze[i][j] = 1;

        if(i == endI && j == endJ) success = 1;

        if(success != 1 && maze[i][j+1] == 0) visit(i, j+1); if(success != 1 && maze[i+1][j] == 0) visit(i+1, j); if(success != 1 && maze[i][j-1] == 0) visit(i, j-1); if(success != 1 && maze[i-1][j] == 0) visit(i-1, j);

        if(success != 1)
            maze[i][j] = 0; return success;

    }