【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 12W........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】

       

【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
     
      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>>m;    for(int i=0;i
       
        >field[i][j];    for(int i=0;i