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.
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。
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.
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。
输出格式
Line 1: The number of ponds in Farmer John‘s field.
一行：水坑的数量
输入输出样例
输入 #1复制
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复制
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.
#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;
}
原文地址：https://www.cnblogs.com/xiongchongwen/p/11820996.html