定義一個二維數組N*M(其中2<=N<=10;2<=M<=10),如5 × 5數組下所示:
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,
};
它表示一個迷宮,其中的1表示牆壁,0表示可以走的路,只能橫着走或豎着走,不能斜着走,要求編程序找出從左上角到右下角的最短路線。入口點爲[0,0],既第一空格是可以走的路。
Input
一個N × M的二維數組,表示一個迷宮。數據保證有唯一解,不考慮有多解的情況,即迷宮只有一條通道。
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:記錄上一個位置
int xcurrent = 0;
int ycurrent = 0;
int count=0;
while(true)
{
if (maze[xcurrent+1][ycurrent]==0 && xcurrent+1<row)
{
//返回上一個位置
if (route[count-1][0]==xcurrent+1 && route[count-1][1]==ycurrent)
{
maze[xcurrent][ycurrent]=1;//設置爲牆
count--;
xcurrent++;
}
else
{
route[count][0]=xcurrent;
route[count][1]=ycurrent;
count++;
xcurrent++;
}
}
else if (maze[xcurrent][ycurrent+1]==0 && ycurrent<line)
{
if (route[count-1][0]==xcurrent && route[count-1][1]==ycurrent+1)
{
maze[xcurrent][ycurrent]=1;//設置爲牆
count--;
ycurrent++;
}
else
{
route[count][0]=xcurrent;
route[count][1]=ycurrent;
count++;
ycurrent++;
}
}
else if (maze[xcurrent-1][ycurrent]==0 && xcurrent-1>=0)
{
if (route[count-1][0]==xcurrent-1 && route[count-1][1]==ycurrent)
{
maze[xcurrent][ycurrent]=1;//設置爲牆
count--;
xcurrent--;
}
else
{
route[count][0]=xcurrent;
route[count][1]=ycurrent;
count++;
xcurrent--;
}
}
else if (maze[xcurrent][ycurrent-1]==0 && ycurrent-1>=0)
{
if (route[count-1][0]==xcurrent && route[count-1][1]==ycurrent-1)
{
maze[xcurrent][ycurrent]=1;//設置爲牆
count--;
ycurrent--;
}
else
{
route[count][0]=xcurrent;
route[count][1]=ycurrent;
count++;
ycurrent--;
}
}
if (xcurrent==row-1 && ycurrent==line-1)
{
route[count][0]=xcurrent;
route[count][1]=ycurrent;
count++;
break;
}
}
for (int i=0;i<count;i++)
{
printf("(%d,%d)\n",route[i][0],route[i][1]);
}
return 0;
}