BFS解决POJ 2386

Description

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 

* 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.

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

#include "iostream"
#include "string"
using namespace std;
int used[105][105];
char map[105][105];
int direction[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
struct point
{
	int x;
	int y;
}queue[10005];
int si, sj, iColumn, iRow, count;
void bfs()
{
	int head = 0, tail = 0;
	used[si][sj] = 1;
	queue[head].x = si;  //搜索的初始x坐标
	queue[head].y = sj;  //搜索的初始y坐标
	head++;
	while(tail<head)
	{
		point temp1 = queue[tail];
		tail++;
		int k, m;
		for(k=0; k<8; k++)
		{
			point temp2;
			temp2.x = temp1.x+direction[k][0];  
			temp2.y = temp1.y+direction[k][1]; 
			if(temp2.x>=0 && temp2.x<iRow && temp2.y>=0 && temp2.y<iColumn && !used[temp2.x][temp2.y] && map[temp2.x][temp2.y]=='W')
			{
				used[temp2.x][temp2.y] = 1;
				queue[head] = temp2;
				head++;
			}
		}
	}
}
int main()
{
	int i, j;
	while(cin>>iRow>>iColumn)
	{
		memset(used, 0, sizeof(used));
		count=0;
		for(i=0; i<iRow; i++)
			for(j=0; j<iColumn; j++)
				cin>>map[i][j];
       　　　　  for(i=0; i<iRow; i++)
			for(j=0; j<iColumn; j++)
			{
				if(map[i][j]=='W' && used[i][j]==0)  //一旦遇到了W，并且没有被访问过就bfs(),count++;
				{
					si = i;
					sj = j;
					bfs();
					count++;
				}
			}
		cout<<count<<endl;
	}
	return 0;
}