華為面試題：迷宮問題 C語言源碼
來源：程序員人生 發(fā)布時(shí)間：2015-04-08 08:43:50 閱讀次數(shù)：4321次
定義1個(gè)2維數(shù)組N*M（其中2<=N<=10;2<=M<=10），如5 × 5數(shù)組下所示：
int maze[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,
};
它表示1個(gè)迷宮，其中的1表示墻壁，0表示可以走的路，只能橫著走或豎著走，不能斜著走，要求編程序找出從左上角到右下角的最短線路。入口點(diǎn)為[0,0],既第1空格是可以走的路。
Input
1個(gè)N × M的2維數(shù)組，表示1個(gè)迷宮。數(shù)據(jù)保證有唯1解,不斟酌有多解的情況，即迷宮只有1條通道。
Output
左上角到右下角的最短路徑，格式如樣例所示。
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
輸出
(0,0)
(1,0)
(2,0)
(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]);
}
}
//走迷宮
//堆棧:記錄上1個(gè)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個(gè)位置
if (route[count⑴][0]==xcurrent+1 && route[count⑴][1]==ycurrent)
{
maze[xcurrent][ycurrent]=1;//設(shè)置為墻
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;//設(shè)置為墻
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;//設(shè)置為墻
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;//設(shè)置為墻
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;
}
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