POJ No.2386【B007】

【B007】Lake Counting【难度B】——————————————————————————————————————————

【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

【Hint】

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

【Source】

USACO 2004 November

【分析】

不要被那么一大串英文吓到，这就是一个统计八连快。。。。。。果断DFS，秒A。。。。。。

【代码】

#include<iostream>
using namespace std;
const int maxn=101;
const int maxm=101;
int n,m;
char field[maxn][maxm];
void dfs(int x,int y)
{
    field[x][y]='.';
    for(int dx=-1;dx<=1;dx++)
    {
        for(int dy=-1;dy<=1;dy++)
        {
        	int nx=x+dx,ny=y+dy;
        	if(0<=nx && nx<n && 0<=ny && ny<m && field[nx][ny]=='W') dfs(nx,ny);
		}
	}
	return ;
}
int main()
{
    int res=0;
    cin>>n>>m;
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            cin>>field[i][j];
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(field[i][j]=='W')
            {
		        dfs(i,j);
				res++;
			}
		}
    }
    cout<<res;
	return 0;
}