Lake Counting

题号：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 

* 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.

 

 

分析：题目给出一个地图，W代表水坑，.代表干地，要你求图中共有多少个大水坑（连在一起的小水坑算一个，周围的九个方位都算），差不多就是一个DFS的模板题，从找的第一个W出发，寻找周围有的‘W'，找到就让它为’ .'，不然会重复操作；

 

 

 

AC代码：

#include<iostream>
#include<cstdio>
#define N 125
using namespace std;
char a[N][N];
int row,col;
void dfs(int x,int y)
{
    a[x][y]='.';
    for (int dx=-1;dx<=1;dx++)
      for (int dy=-1;dy<=1;dy++)
    {
        int nx=x+dx,ny=y+dy;
        if (nx>=0&&nx<=row&&ny>=0&&ny<=col&&a[nx][ny]=='W')
            dfs(nx,ny);
    }
}
int main()
{

    cin>> row>> col;
    for (int i = 0; i <row; i++)
        for (int j = 0; j <col; j++)
            cin >>a[i][j];
    int res = 0;
    for(int i=0;i<row;i++)
        for(int j=0;j<col;j++)
            if (a[i][j] == 'W')//从有积水的地方开始dfs
            {
                dfs(i, j);
                res++;
            }
    cout << res << endl;
    return 0;
}