ZOJ 题目2100 Seeding（DFS）

标签：

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

水水~~

ac代码

#include<stdio.h>
#include<string.h>
int n,m,flag,ans;
char map[10][10];
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int jud(int x,int y)
{
	if(x<0||x>=n)
		return 0;
	if(y<0||y>=m)
		return 0;
	if(map[x][y]=='S')
		return 0;
	return 1;
}
void dfs(int x,int y)
{
	int i;
	//vis[x][y]=1;
	if(ans==0)
	{
		flag=1;
		return;
	}
	if(flag)
		return;
	for(i=0;i<4;i++)
	{
		int tx=x+dx[i];
		int ty=y+dy[i];
		if(jud(tx,ty))
		{
			map[tx][ty]='S';
			ans--;
			dfs(tx,ty);
			map[tx][ty]='.';
			ans++;
		}
	}
}
int main()
{
//	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF,n||m)
	{
		int i,j;
		ans=n*m;
		flag=0;
		for(i=0;i<n;i++)
		{
			scanf("%s",map[i]);
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='S')
					ans--;
			}
		}
		ans--;
		map[0][0]='S';
		dfs(0,0);
		if(flag)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}

ZOJ 题目2100 Seeding（DFS）

标签：

原文地址：http://blog.csdn.net/yu_ch_sh/article/details/47253955