华为面试题：迷宫问题 C语言源码

定义1个2维数组N*M（其中2<=N<=10;2<=M<=10），如5 × 5数组下所示：

int maze[5][5] = {

        0,1,

        0,

};

它表示1个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短线路。入口点为[0,0],既第1空格是可以走的路。

Input

1个N × M的2维数组，表示1个迷宫。数据保证有唯1解,不斟酌有多解的情况，即迷宫只有1条通道。

Output

左上角到右下角的最短路径，格式如样例所示。

5 5
0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0
输出
(0,0)
(1,0)
(2,1)
(2,2)
(2,3)
(2,4)
(3,4)
(4,4)

#include stdio.h #include stdlib.h #include string.h #define MAX_PATH 256 int maze[10][10] = {0}; int route[100][2] = {0}; int main() { int row=0,line=0; scanf(%d %d,&row,&line); for (int i=0;i<row;i++) { for (int j=0;j<line;j++) { scanf(%d,&maze[i][j]); } } //走迷宫 //堆栈:记录上1个位置 int xcurrent = 0; int ycurrent = 0; int count=0; while(true) { if (maze[xcurrent+1][ycurrent]==0 && xcurrent+1<row) { //返回上1个位置 if (route[count⑴][0]==xcurrent+1 && route[count⑴][1]==ycurrent) { maze[xcurrent][ycurrent]=1;//设置为墙 count--; xcurrent++; } else { route[count][0]=xcurrent; route[count][1]=ycurrent; count++; xcurrent++; } } else if (maze[xcurrent][ycurrent+1]==0 && ycurrent<line) { if (route[count⑴][0]==xcurrent && route[count⑴][1]==ycurrent+1) { maze[xcurrent][ycurrent]=1;//设置为墙 count--; ycurrent++; } else { route[count][0]=xcurrent; route[count][1]=ycurrent; count++; ycurrent++; } } else if (maze[xcurrent⑴][ycurrent]==0 && xcurrent⑴>=0) { if (route[count⑴][0]==xcurrent⑴ && route[count⑴][1]==ycurrent) { maze[xcurrent][ycurrent]=1;//设置为墙 count--; xcurrent--; } else { route[count][0]=xcurrent; route[count][1]=ycurrent; count++; xcurrent--; } } else if (maze[xcurrent][ycurrent⑴]==0 && ycurrent⑴>=0) { if (route[count⑴][0]==xcurrent && route[count⑴][1]==ycurrent⑴) { maze[xcurrent][ycurrent]=1;//设置为墙 count--; ycurrent--; } else { route[count][0]=xcurrent; route[count][1]=ycurrent; count++; ycurrent--; } } if (xcurrent==row⑴ && ycurrent==line⑴) { route[count][0]=xcurrent; route[count][1]=ycurrent; count++; break; } } for (int i=0;i<count;i++) { printf((%d,%d) ,route[i][0],route[i][1]); } return 0; }