D - Lake Counting

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.

搜索水题，只要往四周一直搜，搜过的改变值就可以，能搜几次就是几个；

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define pb push_back
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod=1e12+100;
const db e=exp(1);
using namespace std;
const double pi=acos(-1.0);

char a[105][105];

bool dfs(int x,int y)
{   
//  cout<<x<<" "<<y<<endl;

    if(a[x][y]!='W') return false;
    a[x][y]='a';//把经过的位置改变 
    
    
    if(a[x-1][y-1]=='W')//左上 
        dfs(x-1,y-1);
    if(a[x-1][y+1]=='W')//右上 
        dfs(x-1,y+1);
    if(a[x+1][y-1]=='W')//左下 
        dfs(x+1,y-1);   
    if(a[x+1][y+1]=='W')//右下 
        dfs(x+1,y+1);   
    if(a[x-1][y]=='W')//上 
        dfs(x-1,y);
    if(a[x][y-1]=='W')//左 
        dfs(x,y-1);
    if(a[x][y+1]=='W')//右 
        dfs(x,y+1);
    if(a[x+1][y]=='W')//下 
        dfs(x+1,y);
        return true;
}
int solve(int n,int m)
{
    int sum=0;
    rep(i,1,n+1)
    {
        rep(j,1,m+1)
        if(dfs(i,j))
        sum++;
    }
    return sum;
}
int main()
{
        int n,m;
        sf("%d%d%d%d",&n,&m);
        mm(a,'.');
        rep(i,1,n+1)
        {
            sf("%s",&a[i][1]);
            //pf("1%s\n",&d[i]);
            a[i][m+1]='.';
        }
        pf("%d\n",solve(n,m));
}