POJ - 2386：Lake Counting

Lake Counting

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题目

Due to recent rains, water has pooled in various places in Farmer John’s field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water (‘W’) or dry land (’.’). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John’s field, determine how many ponds he has.

输入

• Line 1: Two space-separated integers: N and M

• Lines 2…N+1: M characters per line representing one row of Farmer John’s field. Each character is either ‘W’ or ‘.’. The characters do not have spaces between them.

输出

• Line 1: The number of ponds in Farmer John’s field.

输入样例

10 12
W…WW.
.WWW…WWW
…WW…WW.
…WW.
…W…
…W…W…
.W.W…WW.
W.W.W…W.
.W.W…W.
…W…W.

输出样例

3

提示

OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.

参考代码

#include
const int maxn=100+10;
char arr[maxn][maxn];
int vis[maxn][maxn];
int n,m; 
void dfs(int x,int y)
{
	vis[x][y]=1;//或者可以令arr[i][j]='.'; 
	for(int i=-1;i<=1;i++)
	{
		for(int j=-1;j<=1;j++)
		{
			int nx=x+i,ny=y+j;
			if(0<=nx && nx<n && 0<=ny && ny<m && arr[nx][ny]=='W' && !vis[nx][ny])
				dfs(nx,ny);
		}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	int i,j;
	for(i=0;i<n;i++)
		scanf("%s",arr[i]);
	int cnt=0;
	for(i=0;i<n;i++)
	{
		for(j=0;j<m;j++)
		{
			if(arr[i][j]=='W' && !vis[i][j])
			{
				dfs(i,j);
				cnt++;
			}
		}
	}
	printf("%d\n",cnt);
	return 0;
}