POJ 2386 Lake Counting_steven 解题心得

原题:

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.

 

 

 

我的代码：

#include<iostream>
#include<cstdio>

using namespace std;
int flag = 1;
char ground[110][110];
int tree[110][110];
int square = 0;
int directionX[8] = {1 ,  1 ,  1 ,   0,    0 ,    -1,   -1 , -1 };
int directionY[8] = { 1,  0 ,  -1,   1,    -1,    1,    0,   -1 };
int n, m;

void dfs(int i,int j ,int sq)
{
    if (ground[i][j] == 'W'&&tree[i][j] == 0)
    {
        tree[i][j] = sq;
        //拓展
        for (int k = 0; k < 8; k++)
        {
            if (i + directionX[k] >= n || j + directionY[k] >= m ||
                i + directionX[k] < 0 || j + directionY[k] < 0)            //越界
                    {continue;}
            else
            {
                dfs(i + directionX[k], j + directionY[k], sq);
            }
        }
    }
    return;
}


int main()
{
    
    cin >> n>>m;
    for (int i = 0; i < n; i++)
        scanf("%s", ground[i]);
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (ground[i][j] == 'W'&&tree[i][j] == 0)
            {
                dfs(i, j, ++square);
            }
        }
    }
    cout << square << endl;

    return 0;
}