POJ 2386 Lake Counting 八方向棋盘搜索

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 53301		Accepted: 26062

Description

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代表水塘，. 代表土地，问一共有多少个联通的水塘

 

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
int dir[8][2]={{0,-1},{0,1},{1,0},{-1,0},{-1,1},{1,1},{-1,-1},{1,-1}};
char a[105][105];
int n,m;
int check(int x,int y)
{
    if(x>=0&&x<n&&y>=0&&y<m&&a[x][y]=='W')
        return 1;
    else
        return 0;
}
void dfs(int x,int y)
{
    if(check(x,y)==0)
        return;
    if(a[x][y]=='W')
        a[x][y]='.';
    for(int i=0;i<8;i++)
    {
        int dx,dy;
        dx=x+dir[i][0];
        dy=y+dir[i][1];
        dfs(dx,dy);
    }
}
int main()
{
    cin>>n>>m;
    int cnt=0;
    for(int i=0;i<n;i++)
        cin>>a[i];
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(a[i][j]=='W')
            {
                dfs(i,j);
                cnt++;
            }
        }
    }
    cout<<cnt<<endl;
    return 0;
}