Lake Counting (DFS )

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.

题意分析：给你一块地，W表示有水的地方，‘.’表示地是干的，若一个W与周围的W紧挨着则他们是一整块池塘，问你一共有多少块池塘

思路：从第一行第一列开始扫，遇到W就进入dfs, dfs怎么实现呢？  定义八个方向，上下左右，左上左下，右上右下，表示要搜索的地方，若越界了就continue，否则就判断是否下一块是不是W，若是则将他清除（置为'.'）dfs扫完了就return，找到的数量cnt++,返回到二维数组中去，继续扫。

AC代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<cstdlib>#include<queue>#include<set>#include<map>#include<vector>#include<stack>using namespace std;#define ll long longchar field[105][105];int dir[8][2]={{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};struct NODE{    int x,y;};int n,m;void dfs(int i,int j){    NODE first;    first.x=i;    first.y=j;    queue<NODE> q;    q.push(first);    while(!q.empty()){        first=q.front();        q.pop();        for(int i=0;i<8;i++){            NODE tmp;            tmp.x=first.x+dir[i][0];            tmp.y=first.y+dir[i][1];            if(tmp.x<0||tmp.y<0||tmp.x>=n||tmp.y>=m)continue;            if(field[tmp.x][tmp.y]=='W'){                q.push(tmp);                field[tmp.x][tmp.y]='.';            }        }    }    return ;}int main(){    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)        scanf("%s",field[i]);    int cnt=0;    for(int i=0;i<n;i++)        for(int j=0;j<m;j++)        if(field[i][j]=='W'){            cnt++;            dfs(i,j);        }    printf("%d\n",cnt);    return 0;}