poj 2386 Lake Counting 简单深搜

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本博客介绍了一种方法来计算园子中积水的总面积，并通过积水区域的连通性来确定水洼的数量。该方法适用于园子积水后的水量分析。

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

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

大意：有一个大小为N*M的园子，雨后积了很多水。八连通的积水被认为是在一起的。请求出园子里共有多少个水洼？

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

char a[101][101];
int s,t;
void ss(int x,int y)
{
     if(a[x][y]=='.'|| x<0||x>=s||y<0||y>=t)
        return;
    a[x][y]='.';
    ss(x-1,y-1);
    ss(x-1,y);
    ss(x-1,y+1);
    ss(x,y-1);
    ss(x,y+1);
    ss(x+1,y-1);
    ss(x+1,y);
    ss(x+1,y+1);
}
int main()
{
    while(cin>>s>>t)
    {
        for(int i=0; i<s; i++)
            for(int j=0; j<t; j++)
            {
                cin>>a[i][j];
            }
        int sum=0;
        for(int i=0; i<s; i++)
            for(int j=0; j<t; j++)
            {
                if(a[i][j]=='W')
                {
                    sum++;
                    ss(i,j);
                }
            }
        cout<<sum<<endl;
    }
    return 0;
}