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Lake Counting

发布日期：2021-08-21 13:18:13 浏览次数：48 分类：技术文章

本文共 1685 字，大约阅读时间需要 5 分钟。

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 12W........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.

这题就是让你找有几个连通域，这题是向8个方向发散

#include
       
        #include
        
         #include
         
          using namespace std;char a[110][110];int n,m;int s=0;void bfs(int i,int j){    a[i][j]='.';    if(j+1
          
           =0&&j+1
           
            =0&&a[i-1][j]=='W')        bfs(i-1,j);    if(j-1>=0&&a[i][j-1]=='W')        bfs(i,j-1);    if(i-1>=0&&j-1>=0&&a[i-1][j-1]=='W')        bfs(i-1,j-1);    if(i+1
            
             =0&&a[i+1][j-1]=='W') bfs(i+1,j-1);}int main(){ scanf("%d%d",&n,&m); char c; scanf("%c",&c); for(int i=0;i

转载于:https://www.cnblogs.com/xzxj/p/7236220.html

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