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

3

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

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　　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：

 1 #include <iostream>

 2 

 3 using namespace std;  4 int n,m;  5 char a[105][105];  6 int dx[8] = {0,1,1,1,0,-1,-1,-1};   //八个方向

 7 int dy[8] = {1,1,0,-1,-1,-1,0,1};  8 bool judge(int x,int y)  9 { 10     if(x<1 || x>n || y<1 || y>m) 11         return 1; 12     if(a[x][y]!='W') 13         return 1; 14     return 0; 15 } 16 void dfs(int x,int y) 17 { 18     a[x][y] = '.';  //将‘W’转化为‘.’

19     for(int i=0;i<8;i++){ 20         int nx = x + dx[i]; 21         int ny = y + dy[i]; 22         //如果这一步是‘W’，且没有越界，可以走。

23         if(judge(nx,ny)) 24             continue; 25  dfs(nx,ny); 26  } 27 } 28 int main() 29 { 30     while(cin>>n>>m){ 31         int sum = 0; 32         for(int i=1;i<=n;i++) 33             for(int j=1;j<=m;j++) 34                 cin>>a[i][j]; 35         for(int i=1;i<=n;i++) 36             for(int j=1;j<=m;j++) 37                 if(a[i][j]=='W'){ 38                     sum++; 39  dfs(i,j); 40  } 41         cout<<sum<<endl; 42  } 43     return 0; 44 } 45     

 

Freecode : www.cnblogs.com/yym2013