题目描述

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.

输入格式

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.

输出格式

Line 1: The number of ponds in Farmer John's field.

输入输出样例 #1

输入 #1

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

 输出 #1

3

说明/提示

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

代码为：

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 110;
int n, m;
int dx[] = {1, 1, 1, 0,0,-1,-1,-1};//x轴方向数组
int dy[] = {-1,0,1,1,-1,1,0,-1};//y轴方向数组
char g[N][N];//存储地图
bool st[N][N];//标记是否被淹过,默认为false
int res = 0;//记录有多少个水坑
void dfs(int x,int y)
{

    for (int i = 0; i < 8;i++)
    {
        int a = x + dx[i];
        int b = y + dy[i];

        if(a<0||a>=n||b<0||b>=m)//越界
            continue;
        if(g[a][b]!='W')//不是水
            continue;
        if(st[a][b])//已经被淹过
            continue;
        //标记已被淹过
        st[a][b] = true;//标记已被淹过
        dfs(a, b);//递归
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n;i++)
    cin>>g[i];//输入地图

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (g[i][j] == 'W'&&!st[i][j])//如果是水且未被淹过
            {
                dfs(i, j);//遍历地图
                res++;//水坑数加一
            }
        }
    }
    cout << res;
    return 0;
}