G - Lake Counting

题目描述：

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.
题意：W是水，.是土地。求给出的地图中有几个池塘。注意: 斜方向的连接，也表示这两团水是连接的。
分析：这里也是一个搜索问题。我们可以遍历时如果碰到W，就立马执行dfs，并计数加1。到最后执行了几次dfs，就是几个池塘。

#include"stdio.h"
#include"string.h"
char Map[100][100];
int cont;
int n,k;
//dfs的目的是将W与之相连的W全部用. 替换
void dfs(int X,int Y)
{
    int x[8]={-1,-1,-1,0,0,1,1,1};//主意有8个方向
    int y[8]={-1,0,1,-1,1,-1,0,1};
    for(int i=0;i<8;i++)
    {
        int a=X+x[i];
        int b=Y+y[i];
        if(a>=0&&a=0&&b