Lake Counting

 

问题描述

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

* 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

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

样例输入

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

样例输出

3

很经典的搜索问题，每找到一个water位’W'，就进行dfs把它及它相连的water填充成 ‘.‘，将当前找到的lack抹去，防止重复计算lack。

#include <iostream>
using namespace std;
char map[105][105];
int dx[8]={-1,0,1,0,1,-1,-1,1},
    dy[8]={0,-1,0,1,1,-1,1,-1};
void dfs(int x,int y);
int count=0,m,n;
int main() {
    cin>>n>>m;
    for(int i=0;i<n;i++) {
        for (int j = 0; j < m; j++){
            cin >> map[i][j];
        }
    }
    for(int i=0;i<n;i++) {
        for (int j = 0; j < m; j++){
            if(map[i][j]=='W') {
                count++;
                dfs(i,j);
            }
        }
    }
    cout<<count<<endl;
    return 0;
}
void dfs(int x,int y)
{
    map[x][y]='.';

    for(int i=0;i<8;i++){
        int px=x+dx[i],py=y+dy[i];
        if(px>=0&&px<n&&py>=0&&py<m&&map[px][py]=='W')
            dfs(px,py);
    }
}