Question

高分题～～C语言高手请进。。迷宫问题

把一下程序的输出结果改为。。走过的路径为2（时间紧迫，尽快。答对有分追加）
即把
1,1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,1,1,
1,0,1,1,1,0,0,1,0,1,
1,0,0,0,1,0,0,0,0,1,
1,0,1,0,0,0,1,0,1,1,
1,0,1,1,1,1,0,0,1,1,
1,1,1,0,0,0,1,0,0,1,
1,1,1,1,1,1,1,1,1,1,
改为
1,1,1,1,1,1,1,1,1,1,
1,2,0,1,0,0,0,1,0,1,
1,2,0,1,0,0,0,1,0,1,
1,2,0,0,0,1,1,0,1,1,
1,2,1,1,1,0,0,1,0,1,
1,2,2,2,1,2,2,2,0,1,
1,0,1,2,2,2,1,2,1,1,
1,0,1,1,1,1,0,2,1,1,
1,1,1,0,0,0,1,2,2,1,
1,1,1,1,1,1,1,1,1,1,

程序代码如下：

#include<stdio.h>
#define N1 10
#define N2 10
int getpath(int maze[N1][N2])//在迷宫中探路的函数
{
int stack[N1*N2][2];
int i,x=1,y=1,ok,top=0;//可通路径上第一个点（入口点）的坐标进入路径数组stack
stack[top][0]=x;
stack[top][1]=y;//循环探路。加1则向前，减1则退回一步

while(1)
{
ok=0;//标志能否向前
if (maze[x+1][y]==0)
{x=x+1;ok=1;}//向下
else
if (maze[x][y+1]==0)
{y=y+1;ok=1;}//向右
else
if (maze[x][y-1]==0)
{y=y-1;ok=1;}//向左
else
if (maze[x-1][y]==0)
{x=x-1;ok=1;}//向上
if (!ok)//4方向均不通
{ top--;
if (top==0)//栈空，无出路
{printf("迷宫无出路\n");
return 0; }
x=stack[top][0];//回退后路径上最后一个点的X坐标存入X
y=stack[top][1];}//回退后路径上最后一个点的Y坐标存入Y
else//若能走通
{
maze[x][y]=2;
top++;
stack[top][0]=x;//新进入的点的X坐标进入路径数组
stack[top][1]=y;//新进入的点的Y坐标进入路径数组

if (x==N1-2&&y==N2-2)//到达出口
{printf("路径为：\n");
for(i=0;i<top;i++)
{printf("(%d,%d)-->",stack[i][0],stack[i][1]);
if((i+1)%5==0)
printf("\n");
}
printf("(%d,%d)\n",stack[top][0],stack[top][1]);
return 1;
}
}
}
}

void printmaze(maze)//输出迷宫数组函数
int maze[N1][N2];
{
int i,j;
printf("迷宫为:\n");
for (i=0;i<N1;i++)
{
for (j=0;j<N2;j++)
printf("%2d",maze[i][j]);
printf("\n");
}
}

void main()
{
int A[N1][N2]=
{
1,1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,0,1,
1,0,1,1,1,0,0,0,0,1,
1,0,0,0,1,0,0,0,0,1,
1,0,1,0,0,0,1,0,0,1,
1,0,1,1,1,0,1,1,0,1,
1,1,0,0,0,0,0,0,0,1,
1,1,1,1,1,1,1,1,1,1,

}; //The maze

printmaze(A);
getpath(A);
}

Answer 1

改成这样就可以了
没有什么大问题,就是printmaze及其参数定义有点问题.

#include<stdio.h>
#define N1 10
#define N2 10
int maze[N1][N2];
int getpath(int maze[N1][N2])//在迷宫中探路的函数
{
int stack[N1*N2][2];
int i,x=1,y=1,ok,top=0;//可通路径上第一个点（入口点）的坐标进入路径数组stack
stack[top][0]=x;
stack[top][1]=y;//循环探路。加1则向前，减1则退回一步

while(1)
{
ok=0;//标志能否向前
if (maze[x+1][y]==0)
{
x=x+1;
ok=1;
}//向下
else if (maze[x][y+1]==0)
{
y=y+1;
ok=1;
}//向右
else if (maze[x][y-1]==0)
{
y=y-1;
ok=1;
}//向左
else if (maze[x-1][y]==0)
{
x=x-1;ok=1;
}//向上
if (!ok)//4方向均不通
{
top--;
if (top==0)//栈空，无出路
{
printf("迷宫无出路\n");
return 0;
}
x=stack[top][0];//回退后路径上最后一个点的X坐标存入X
y=stack[top][1];
}//回退后路径上最后一个点的Y坐标存入Y
else//若能走通
{
maze[x][y]=2;
top++;
stack[top][0]=x;//新进入的点的X坐标进入路径数组
stack[top][1]=y;//新进入的点的Y坐标进入路径数组

if (x==N1-2&&y==N2-2)//到达出口
{
printf("路径为：\n");
for(i=0;i<top;i++)
{
printf("(%d,%d)-->",stack[i][0],stack[i][1]);
if((i+1)%5==0)
printf("\n");
}
printf("(%d,%d)\n",stack[top][0],stack[top][1]);
return 1;
}
}
}
}

void printmaze(int maze[N1][N2])//输出迷宫数组函数

{
int i,j;
printf("迷宫为:\n");
for (i=0;i<N1;i++)
{
for (j=0;j<N2;j++)
printf("%2d",maze[i][j]);
printf("\n");
}
}

main()
{
int A[N1][N2]=
{
1,1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,0,1,
1,0,1,1,1,0,0,0,0,1,
1,0,0,0,1,0,0,0,0,1,
1,0,1,0,0,0,1,0,0,1,
1,0,1,1,1,0,1,1,0,1,
1,1,0,0,0,0,0,0,0,1,
1,1,1,1,1,1,1,1,1,1,

}; //The maze

printmaze(A);
getpath(A);
printmaze(A);
getchar();
}

Answer 2

用的笨法子。。。
#include<stdio.h>
#define N1 10
#define N2 10

void printmaze(int maze[N1][N2])//输出迷宫数组函数

{
int i,j;

for (i=0;i<N1;i++)
{
for (j=0;j<N2;j++)
printf("%2d",maze[i][j]);
printf("\n");
}
}

int getpath(int maze[N1][N2])//在迷宫中探路的函数
{
int stack[N1*N2][2];
int i,m,n,x=1,y=1,ok,top=0;//可通路径上第一个点（入口点）的坐标进入路径数组stack
stack[top][0]=x;
stack[top][1]=y;//循环探路。加1则向前，减1则退回一步

while(1)
{
ok=0;//标志能否向前
if (maze[x+1][y]==0)
{x=x+1;ok=1;}//向下
else
if (maze[x][y+1]==0)
{y=y+1;ok=1;}//向右
else
if (maze[x][y-1]==0)
{y=y-1;ok=1;}//向左
else
if (maze[x-1][y]==0)
{x=x-1;ok=1;}//向上
if (!ok)//4方向均不通
{ top--;
if (top==0)//栈空，无出路
{printf("迷宫无出路\n");
return 0; }
x=stack[top][0];//回退后路径上最后一个点的X坐标存入X
y=stack[top][1];}//回退后路径上最后一个点的Y坐标存入Y
else//若能走通
{
maze[x][y]=2;
top++;
stack[top][0]=x;//新进入的点的X坐标进入路径数组
stack[top][1]=y;//新进入的点的Y坐标进入路径数组

if (x==N1-2&&y==N2-2)//到达出口
{
for(m=1;m<N1;m++)
for(n=1;n<N2;n++)
if(maze[m][n]==2)maze[m][n]=0;
printf("结果为：\n");
for(i=0;i<top;i++)
{
maze[stack[i][0]][stack[i][1]]=2;

}
maze[N1-2][N2-2]=2;
printmaze(maze);
return 1;
}
}
}
}

int main()
{
int A[N1][N2]=
{
1,1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,1,1,
1,0,1,1,1,0,0,1,0,1,
1,0,0,0,1,0,0,0,0,1,
1,0,1,0,0,0,1,0,1,1,
1,0,1,1,1,1,0,0,1,1,
1,1,1,0,0,0,1,0,0,1,
1,1,1,1,1,1,1,1,1,1

}; //The maze
printf("迷宫为:\n");
printmaze(A);
getpath(A);
getchar();
}