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
Author: ZHANG, Zheng
Source: Zhejiang University Local Contest 2004
这道题刚开始写好后,运行一下没什么问题,但是提交之后就是WA,找了半天也没找到错误,于是就把代码删了重新写了一遍,然后就AC了。。。。
这道题主要的地方就是一个回溯问题,即如果此路不通就把石头重新变回土地,最开始学时懵懵懂懂,并且也没有深入去理解,不过现在理解起来容易多了
题意:石头"S"不能过,用过的土地"."不能再用,能否把土地一次性种上种子
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int i,j,m,n,a,flag,k;
char map[10][10];
void dfs(int x,int y)
{
if(k==n*m)
{
flag=1;
return ;
}
if(x<0||x>=m||y<0||y>=n||map[x][y]=='S')
return ;
if(flag)
return ;
a++;
if(a>1500)//这是个很值得学习的地方,如果查找的次数,超过了这个,就不用再继续下去了,节省了代码运行时间
return ;
map[x][y]='S';
k++;
dfs(x+1,y);
dfs(x-1,y);
dfs(x,y+1);
dfs(x,y-1);
k--;
map[x][y]='.';
}
int main()
{
while(scanf("%d%d",&m,&n),m+n)
{
a=0;
flag=0;
k=0;
for(i=0;i<m;i++)
scanf("%s",map[i]);
for(i=0;i<m;i++)
for(j=0;j<n;j++)
if(map[i][j]=='S')
k++;
dfs(0,0);
if(flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}