poj 2386:Lake Counting（简单DFS深搜）

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18201		Accepted: 9192

Description

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.

Input

* 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.

Output

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

Sample Input

10 12

W........WW.

.WWW.....WWW

....WW...WW.

.........WW.

.........W..

..W......W..

.W.W.....WW.

W.W.W.....W.

.W.W......W.

..W.......W.

Sample Output

Hint

OUTPUT DETAILS:

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

Source

USACO 2004 November

　　简单深搜。

　　遍历迷宫中所有的点，如果是‘W’，开始dfs搜索，将它临近的八个方向是‘W’的点全部走遍，并将走过的点变为‘.’，这样这一块‘W’区域就全部走完。一共走过了多少个这样的区域，就是结果。

　　代码：

 #include <iostream>

 using namespace std;

 int n,m;

 char a[][];

 int dx[] = {,,,,,-,-,-};   //八个方向

 int dy[] = {,,,-,-,-,,};

 bool judge(int x,int y)

 {

     if(x< || x>n || y< || y>m)

         return ;

     if(a[x][y]!='W')

         return ;

     return ;

 }

 void dfs(int x,int y)

 {

     a[x][y] = '.';  //将‘W’转化为‘.’

     for(int i=;i<;i++){

         int nx = x + dx[i];

         int ny = y + dy[i];

         //如果这一步是‘W’，且没有越界，可以走。

         if(judge(nx,ny))

             continue;

         dfs(nx,ny);

     }

 }

 int main()

 {

     while(cin>>n>>m){

         int sum = ;

         for(int i=;i<=n;i++)

             for(int j=;j<=m;j++)

                 cin>>a[i][j];

         for(int i=;i<=n;i++)

             for(int j=;j<=m;j++)

                 if(a[i][j]=='W'){

                     sum++;

                     dfs(i,j);

                 }

         cout<<sum<<endl;

     }

     return ;

 }

Freecode : www.cnblogs.com/yym2013

巴特西

poj 2386:Lake Counting（简单DFS深搜）

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