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  • POJ 2386 Lake Counting dfs

    Lake Counting
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 37384   Accepted: 18586

    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.
     
     
     
     
     
    思路:扫描任意'W'的8个方向,全部替换成 '.' ,看能扫到几个'W' ,就是结果。
     
     
    代码:
    #include <iostream> 
    using namespace std;
    typedef long long ll;
    
    char a[110][110];
        int m,n;
    void dfs(int x ,int y){
        if(x < 0||x >= n||y < 0||y >= m){
            return;
        }
        a[x][y] = '.';
        
        for(int i = -1;i <= 1; i++){
            for(int j = -1;j <= 1; j++){
                if(a[x+i][y+j] == 'W'){
                    dfs(x+i,y+j);
                }
            }
        }
        return;
    }
    
    
    int main() {
         int ans = 0;
     
      cin >> n >> m;
      for(int i = 0;i < n; i++){
          for(int j = 0;j < m; j++){
              cin >> a[i][j];
            }
        }
        
        for(int i = 0;i < n; i++){
            for(int j = 0;j < m; j++){
                if(a[i][j] == 'W'){
                    dfs(i,j);
                    ans++;
                }
            }
        }
        cout << ans << endl;
      return 0;
    }
    //  writen by zhangjiuding
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  • 原文地址:https://www.cnblogs.com/zhangjiuding/p/7663397.html
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