2019_GDUT_新生专题I选集 B POJ-2386

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.
Input

• Line 1: Two space-separated integers: N and M
• Line 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.
  Output
• Line 1: The number of ponds in Farmer John’s field.
  Sample Input
  10 12
  W…WW.
  .WWW…WWW
  …WW…WW.
  …WW.
  …W…
  …W…W…
  .W.W…WW.
  W.W.W…W.
  .W.W…W.
  …W…W.

Sample Output
3

分析：bfs搜索，遍历整个数组，遇到W就入队，并把与其相连的W全部入队，入队后将该元素标记为’.’，每bfs一次答案就加一。需要注意的是这题把一个点周围八个点全部视为相连的。
由于数据规模较小（100*100），也可以dfs，每次找到一个W就dfs这个W周围是否还有并标记。

代码：
BFS：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,a[110][110],ans;

void bfs(int x1,int y1)
{
	int queue[10100][2],l=0,r=1,x,y;
	int p[]={0,0,1,-1,-1,1,1,-1},q[]={1,-1,0,0,-1,-1,1,1};
	memset(queue,0,sizeof(queue));
	queue[1][1]=x1;
	queue[1][2]=y1;
	a[x1][y1]=0;
	while (l<r)
	{
		l++;
		for (int i=0;i<8;i++)
		{
			x=queue[l][1]+p[i];
			y=queue[l][2]+q[i];
			if (x<=n && y<=m && a[x][y]==1 && x>=1 && y>=1)
			{
				r++;
				queue[r][1]=x;
				queue[r][2]=y;
				a[x][y]=0;
			}
		}
	}
	ans++;
}

int main()
{
	char s[110];
	cin>>n>>m;
	for (int i=1;i<=n;i++)
	{
		scanf("%s",s);
		for (int j=0;j<strlen(s);j++)
			if (s[j]=='W') a[i][j+1]=1;
	}
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<=m;j++)
		{
		 	if (a[i][j])
			bfs(i,j);
		}
	}
	cout<<ans;
	return 0;
}

DFS:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[110][110],n,m,ans=0;
int dx[]={0,0,-1,1,-1,1,-1,1},dy[]={1,-1,0,0,1,1,-1,-1};

void dfs(int x1,int y1)
{
	int x,y;
	for (int i=0;i<8;i++)
	{
		x=x1+dx[i];
		y=y1+dy[i];
		if (x>0 && y>0 && x<=n && y<=m && a[x][y]==1)
		{
			a[x][y]=0;
			dfs(x,y);
		}
	}
}

int main()
{
	memset(a,0,sizeof(a));
	char s[110];
	cin>>n>>m;
	for (int i=1;i<=n;i++)
	{
		cin>>s;
		for (int j=0;j<m;j++)
		{
			if (s[j]=='W')
			a[i][j+1]=1;
		}
	}
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<=m;j++)
		{
			if (a[i][j])
			{
				ans++;
				a[i][j]=0;
				dfs(i,j);
			}
		}
	}
	cout<<ans;
}