P1596 [USACO10OCT]Lake Counting S

 

# [USACO10OCT]Lake Counting S

## 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

## 样例 #1

### 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.
```

### 样例输出 #1

```
3
```

## 提示

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

【思路】

直接套dfs模板， 稍微修改即可通过。

【AC代码】

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<string>
#include<vector>
using namespace std;
const int N=1e3+10;
inline int fread()
{
	char ch=getchar();
	int n=0,m=1;
	while(ch<'0' or ch>'9')
	{
		if(ch=='-')m=-1;
		ch=getchar();
	}
	while(ch>='0' and ch<='9')n=(n<<3)+(n<<1)+ch-48,ch=getchar();
	return n*m;
}
void fwrite(int n)
{
	if(n>9)fwrite(n/10);
	putchar(n%10+'0');
}
int n,m,ans,a[N]={0,-1,-1,-1,0,0,1,1,1},b[N]={0,-1,0,1,-1,1,-1,0,1};
char ch[N][N];
void dfs(int x,int y)
{
	ch[x][y]=='.';//被搜索过
	int z,p;
	for(int i=1;i<9;i++)
	{
		z=x+a[i],p=y+b[i];
		if(z<1 or z>n or p<1 or p>m or ch[z][p]=='.')continue;//判断边界条件
		ch[z][p]='.',dfs(z,p);//符合条件，将ch[z][p]设为被搜索过，继续往下搜
	}
	return;
}
signed main()
{
	n=fread(),m=fread();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)cin>>ch[i][j];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(ch[i][j]=='W')ans++,dfs(i,j);//如果是水，记录水洼个数，并进行搜索
	fwrite(ans);
	return 0;
}

调了半天调不出来干脆换了一种判断方式，然后就过了。 

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