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  • poj 2386 Lake Counting(dfs)

    Lake Counting
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 28220   Accepted: 14160

    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.

    Java AC 代码

    import java.util.Scanner;
    
    public class Main {
        
        static String field[][];
        
        static int rows;
        static int columns;
        
        static int ponds;
        
        static boolean marked[][];
        
        static int dx[] = {-1, 0, 1, -1, 1, -1, 0, 1};
        static int dy[] = {1, 1, 1, 0, 0, -1, -1, -1};
        
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            rows = sc.nextInt();
            columns = sc.nextInt();
            field = new String[rows][columns];
            marked = new boolean[rows][columns];
            for(int i = 0; i < rows; i++) {
                String line = sc.next();
                for(int j = 0; j < columns; j++) {
                    field[i][j] = line.substring(j, j + 1);
                }
            }
            
            for(int i = 0; i < rows; i++)
                for(int j = 0; j < columns; j++) {
                    if(!marked[i][j] && field[i][j].equals("W")) {
                        dfs(i, j);
                        ponds++;
                    }
                }
            System.out.println(ponds);
                
        }
        
        public static void dfs(int row, int col) {
            
            for(int i = 0; i < 8; i++) {
                int _row = row + dy[i];
                int _col = col + dx[i];
                if(_row < rows && _row >= 0 && _col < columns && _col >= 0 && !marked[_row][_col] && field[_row][_col].equals("W")) {
                    marked[_row][_col] = true;
                    dfs(_row, _col);
                }
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/kkkkkk/p/5539882.html
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