Lake Counting

第一部分：题目

题目链接：http://poj.org/problem?id=2386

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.

第二部分：思路

深度优先搜索，讲白点就是每种可能都扫一遍。这题的意思就是找出n*m区域中有多少水洼：就是相连的：上下左右还有对角。那么就可以从左上角开始，遇到W就把它置为'.',然后看它的8个”邻居“是否是W，是的话把它置为‘.’，并且以它为中点再看它的8个邻居。这样整个过程结束就相当于把一个水洼填满了'.'。所以做了多少次就相当于有多少处水洼。

 

第三部分 ：代码

#include<iostream>
using namespace std;
char field[100][100]; 
int count=0;
int n,m;
void dfs(int i,int j)
{
    field[i][j]='.';
    int dx,dy;
    //用循环做8个方向的处理 
    for(dx=-1;dx<=1;dx++)
    {
        for(dy=-1;dy<=1;dy++)
        {
            int x,y;
            x=i+dx;
            y=j+dy;
            if(x>=0&&y>=0&&field[x][y]=='W')
            {
                dfs(x,y);
            }
        }
    }    
}
int main()
{
    int j,i;
    cin>>n>>m;
    for(i=0;i<n;i++)
    {
        for(j=0;j<m;j++)
        {
            cin>>field[i][j];
        }
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<m;j++)
        {
            if(field[i][j]=='W')
            {
                dfs(i,j);
                count++;
            }
        }
    }
    cout<<count<<endl;
    return 0;
}