Seeding
2015-08-09 16:03
246 查看
Problem Description
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
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
#include<stdio.h> char a[7][7]; int n,m,num,flag; int u[4]={0,1,0,-1}; int v[4]={1,0,-1,0}; int can(int x,int y) { if(x<1||x>n) return 0; if(y<1||y>m) return 0; if(a[x][y]=='S') return 0; return 1; } void dfs(int x,int y) { int i; if(num==0) { flag=1;//这里不用flag换成int型返回值为一就会出错 return ; } for(i=0;i<4;i++) { int tx=x+u[i]; int ty=y+v[i]; if(can(tx,ty)) { a[tx][ty]='S'; num--; dfs(tx,ty); a[tx][ty]='.'; num++; } } } int main() { int i,j; while(scanf("%d%d",&n,&m),n) { flag=0; num=n*m; for(i=1;i<=n;i++) { getchar(); for(j=1;j<=m;j++) { scanf("%c",&a[i][j]); if(a[i][j]=='S') num--; } } num--; a[1][1]='S'; dfs(1,1); if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }
相关文章推荐
- activity启动模式区别和优化
- 控制线程顺序执行
- 面试准备--java垃圾回收机制、内存管理
- 音乐盒录像带第二版之注册并取得ControllerEvent方法
- C++ 几种排序算法
- android jni 调用结构体示例
- RunLoop学习笔记
- NOJ2098 3_A+B(III)(大数加法)
- PowerDesigner中Name与Code同步的问题
- 二叉树之打印二叉树两节点路径
- 第23章 尝试互联网(2)
- uva10375 Choose and Divide(唯一分解定理)
- 模拟-FZU-2150-Fire Game
- util.bcel.classfile.ClassFormatException: Invalid byte tag in constant pool: 15
- Win7 Centos7安装双系统
- C++: Vector详解
- SharePreferences——存储数据
- 共享存储函数
- HDOJ1009
- python 标准库-argparse 学习