洛谷P1596 Lake Counting S

题目：

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

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

解析：

额

我已经两个月没有写博客了

毕竟

我是学生呀

前一阵子的考试

让我有点焦头烂额

好吧

这道题

和我两个月前

写的题一样

都是DFS

这道题的思路也十分简单

你需要找出水坑

有多少个

所以就是

你只要遇到水

就要把ans++

同时把联通块去掉

就可以了

下面是代码

代码：

#include
#include
#include
#include
using namespace std;
int xxx[9]={0,-1,-1,-1,0,0,1,1,1};
int yyy[9]={0,-1,0,1,-1,1,-1,0,1};
int n,m,sum=0;
int z[110][110];
void ddd(int x,int y)
{
	z[x][y]=0;
	int xx,yy;
	for(int i=1;i<=8;i++)
	{
		xx=x+xxx[i];yy=y+yyy[i];
    	if(z[xx][yy]==1&&xx>0&&xx<=n&&yy>0&&yy<=m) 
    	ddd(xx,yy);
	}
}
int main()
{
	char ss;
	memset(z,0,sizeof(z));
	cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			cin>>ss;
			if(ss=='W')
			z[i][j]=1;
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(z[i][j]==1) 
			{
				ddd(i,j);
				sum++;
			}
		}
	}
	cout<<sum<<endl;
	return 0;
}