zoj 2100 Seeding

Seeding

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

源代码

<span style="font-size:18px;">#include<stdio.h>#include<string.h>int n,m,k;int sum,count,flag;char map[10][10];void dfs(int x,int y){if(x>0&&x<=n&&y>0&&y<=m&&map[x][y]==0){sum++;map[x][y]=1;if(sum==n*m-count){flag=1;return;} dfs(x+1,y);dfs(x-1,y);dfs(x,y+1);dfs(x,y-1);sum--;//如果各个分支都不符合，则返回原来的节点 map[x][y]=0;}}int main(){while(~scanf("%d%d",&n,&m)&&n){sum=0;flag=0;count=0;for(int i=1;i<=n;i++){getchar();for(int j=1;j<=m;j++){scanf("%c",&k);if(k=='S'){map[i][j]=1;count++;}elsemap[i][j]=0;}}//for(int i=1;i<=n;i++)//for(int j=1;j<=m;j++)//dfs(i,j);dfs(1,1);//注意，不能写成上面那种~因为每次起点都是（1,1） if(flag)printf("YES\n");elseprintf("NO\n");} return 0;} </span>

想了好久想不通，看了同学的代码，思路顿时打开了，毕竟小菜鸟~