P1596 [USACO10OCT]湖计数Lake Counting

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.

由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水(‘W‘) 或是旱地(‘.‘)。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。

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.

第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是‘W‘或‘.‘，它们表示网格图中的一排。字符之间没有空格。

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

一行：水坑的数量

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

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int fxx[9]={0,-1,-1,-1,0,0,1,1,1};//x方向
int fxy[9]={0,-1,0,1,-1,1,-1,0,1};//y方向
int n,m,ans;
char a[101][101];
int read(){
	int a=0,b=1;
	char ch=getchar();
	while((ch<‘0‘||ch>‘9‘)&&(ch!=‘-‘)){
		ch=getchar();
	}
	if(ch==‘-‘){
		b=-1;
		ch=getchar();
	}
	while(ch>=‘0‘&&ch<=‘9‘){
		a=a*10+ch-‘0‘;
		ch=getchar();
	}
	return a*b;
}
void dfs(int x,int y){
    int r,c;
    a[x][y]=‘.‘;
    for (int i=1;i<=8;i++){
        r=x+fxx[i];
        c=y+fxy[i];
        if(r<1||r>n||c<1||c>m||a[r][c]==‘.‘){//判断是否出界
            continue;
		}
        a[r][c]=‘.‘;
        dfs(r,c);
    }
}
int main(){
	n=read(),m=read();
    for (int i=1;i<=n;i++){
        for (int j=1;j<=m;j++){
			cin>>a[i][j];
		}
    }
    ans=0;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=m;j++){
            if (a[i][j]==‘W‘){
                ans++;
                dfs(i,j);
            }
        }
    }
    printf("%d",ans);
    return 0;
}