Poj 2386 Lake counting

Source: http://poj.org/problem? Id = 2386

Lake counting Time limit:1000 ms   Memory limit:65536 K Total submissions:20124   Accepted:10139 Description Due to recent rains, water has pooled in varous 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 wowould like to figure out How sort 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 distriof Farmer John's field, determine how he 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 Usaco 2004 November

Question: Question !!! Two mistakes in one sentence:, how many waters are there in a 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. At first, we understood that two "W" would be considered as one piece of water-and there was another direction. We only considered the four directions of the upper, lower, and lower directions, leading to a long Wa ..

Problem: DFS (it does not feel like a standard DFS and does not need to be traced back )....

AC code:

#include<iostream>#include<string>using namespace std;int dir[8][2]={    {0,1},{0,-1},    {1,0},{-1,0},    {1,1},{1,-1},    {-1,-1},{-1,1}};string map[105];int dx,dy,count=0;bool flag;void dfs(int x,int y){    for(int i=0;i<8;i++){    int tempx=x+dir[i][0],tempy=y+dir[i][1];    if(tempx>=0&&tempx<dx&&tempy>=0&&tempy<dy&&map[tempx][tempy]=='W'){    map[tempx][tempy]='.';    dfs(tempx,tempy);    }    }}int main(){    cin>>dx>>dy;    for(int i=0;i<dx;i++)    cin>>map[i];    for(int i=0;i<dx;i++)    for(int j=0;j<dy;j++){    if(map[i][j]=='W'){    count++;    map[i][j]='.';    dfs(i,j);    }    }    cout<<count<<endl;    return 0;}