为了账号安全,请及时绑定邮箱和手机立即绑定

ZOJ Problem Set - 2100

标签:
产品

ZOJ Problem Set - 2100

Seeding

Time Limit: 1 Second      Memory Limit: 32768 KB

It is spring time and farmers have to plant seeds in the field. Tom has a nice field, which is a rectangle with n * m squares. There are big stones in some of the squares.

Tom has a seeding-machine. At the beginning, the machine lies in the top left corner of the field. After the machine finishes one square, Tom drives it into an adjacent square, and continues seeding. In order to protect the machine, Tom will not drive it into a square that contains stones. It is not allowed to drive the machine into a square that been seeded before, either.

Tom wants to seed all the squares that do not contain stones. Is it possible?


Input

The first line of each test case contains two integers n and m that denote the size of the field. (1 < n, m < 7) The next n lines give the field, each of which contains m characters. 'S' is a square with stones, and '.' is a square without stones.

Input is terminated with two 0's. This case is not to be processed.


Output

For each test case, print "YES" if Tom can make it, or "NO" otherwise.


Sample Input

4 4
.S..
.S..
....
....
4 4
....
...S
....
...S
0 0


Sample Output

YES
NO

=============

典型的DFS题,很不错。

题目很简单,从左上角出发,判断是否能完全把地耕玩(拖拉机不能在石块和已经耕好的地上行驶。)

?

#include <stdio.h>#include <string.h>int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};//四个方向 char map[10][10];//定义一张地图 int done;int Flag;int n,m;void dfs(int x,int y) {  int i;  if(x<1 || y<1 || x>n || y>m)   return;//判断是否越界    done++;   if(done==n*m)//如果成功则返回   {     Flag=1;     return;  }  for(i=0;i<4;i++)  {     if(map[x+dir[i][0]][y+dir[i][1]]=='.')     {               map[x+dir[i][0]][y+dir[i][1]]='S';//标记已经耕过的土地          dfs(x+dir[i][0],y+dir[i][1]);     }  }  done--;  map[x][y]='.';}int main(){   int i,j;     memset(map,'S',sizeof(map));   while(scanf("%d%d%*c",&n,&m)!=EOF)   {      if(n==0 && m==0)      break;      done=0;      Flag=0;      for(i=1;i<=n;i++)      {                   for(j=1;j<=m;j++)         {            scanf("%c",&map[i][j]);            if(map[i][j]=='S')            done++;         }         getchar();             }                map[1][1]='S';      dfs(1,1);      if(Flag==1)      printf("YES\n");      else      printf("NO\n");   }

?

   return 0; }
点击查看更多内容
TA 点赞

若觉得本文不错,就分享一下吧!

评论

作者其他优质文章

正在加载中
  • 推荐
  • 评论
  • 收藏
  • 共同学习,写下你的评论
感谢您的支持,我会继续努力的~
扫码打赏,你说多少就多少
赞赏金额会直接到老师账户
支付方式
打开微信扫一扫,即可进行扫码打赏哦
今天注册有机会得

100积分直接送

付费专栏免费学

大额优惠券免费领

立即参与 放弃机会
意见反馈 帮助中心 APP下载
官方微信

举报

0/150
提交
取消