Coursera Algorithms Programming Assignment 1: Percolation

题目来源http://coursera.cs.princeton.edu/algs4/assignments/percolation.html

作业分为两部分：建立模型和仿真实验。

最关键的部分就是建立模型对象。模型对象要求如下：

The model.  We model a percolation system using an n-by-n grid of sites. Each site is either open or blocked. A full site is an open site that can be connected to an open site in the top row via a chain of neighboring (left, right, up, down) open sites. We say the system percolates if there is a full site in the bottom row. In other words, a system percolates if we fill all open sites connected to the top row and that process fills some open site on the bottom row. (For the insulating/metallic materials example, the open sites correspond to metallic materials, so that a system that percolates has a metallic path from top to bottom, with full sites conducting. For the porous substance example, the open sites correspond to empty space through which water might flow, so that a system that percolates lets water fill open sites, flowing from top to bottom.)

边界要求：  By convention, the row and column indices are integers between 1 and n, where (1, 1) is the upper-left site: Throw a java.lang.IllegalArgumentException if any argument to open(), isOpen(), or isFull() is outside its prescribed range. The constructor should throw a java.lang.IllegalArgumentException if n ≤ 0.

性能要求：  The constructor should take time proportional to n²; all methods should take constant time plus a constant number of calls to the union–find methods union(), find(), connected(), and count().

我的分析

本次作业根据教授在视频课上提示，可以在grid的上方和下方各加入一个虚节点，grid第一行的open节点都与top虚节点连通，grid最后一行的open节点都与bottom虚节点连通。这样只需判断top虚节点与bottom虚节点是否连通就知道grid是否渗透，而不需要去一一选取特定节点比对了。照着这个思路，我实现了下述模型代码，作业得分98。值得注意的是，模型代码的main中测试方法不是仅仅进行各本地测试就可以了，提交作业的时候会进行自动脚本测试，所以提交的版本main方法中必须读取args[0]中的文件名，并加载文件内容进行生成grid和open对应的site。

import edu.princeton.cs.algs4.In;
import edu.princeton.cs.algs4.StdOut;
import edu.princeton.cs.algs4.WeightedQuickUnionUF;

public class Percolation {
	private static final boolean BLOCK = false; // block state
	private static final boolean OPEN = true; // open state

	/* topUF bottomUF n 均为final是因为它们只在构造函数时初始化，后续其值未发生变化 */
	private final WeightedQuickUnionUF topUF; // 用来记录与top虚节点的连通性
	private final WeightedQuickUnionUF bottomUF;// 用来记录与bottom虚节点的连通性
	private final int n;

	private boolean[][] grid;
	private boolean percolateFlag = false; // grid是否渗透的标志
	private int openedNum = 0;// 已经open的site数目

	public Percolation(int n) {
		// create n-by-n grid, with all sites blocked
		if (n < 1)
			throw new IllegalArgumentException("grid size should be bigger than one !");
		this.n = n;
		topUF = new WeightedQuickUnionUF(n * n + 1); // 多了一个节点的空间，位置n*n处用来代表虚节点
		bottomUF = new WeightedQuickUnionUF(n * n + 1); // 多了一个节点的空间，位置n*n处用来代表虚节点
		grid = new boolean[n][n];
		// 初始化grid设为block
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				grid[i][j] = BLOCK;
	}

	private void validate(int row, int col) {
		if (row < 1 || col < 1 || row > n || col > n)
			throw new IllegalArgumentException("input row or col is not illegal!");
	}

	public void open(int row, int col) {
		// open site (row, col) if it is not open already
		validate(row, col);
		if (grid[row - 1][col - 1] == OPEN)
			return;

		grid[row - 1][col - 1] = OPEN;
		openedNum++;

		// n为1时，open一个节点就达到渗透要求
		if (n == 1) {
			topUF.union(0, 1);
			bottomUF.union(0, 1);
			percolateFlag = true;
			return;
		}

		// 第一行的所有节点都与top虚节点连通
		if (row == 1)
			topUF.union(n * n, col - 1);

		// 最后一行的所有节点都与bottom虚节点连通
		if (row == n)
			bottomUF.union(n * n, (n - 1) * n + col - 1);

		// 与上方节点的连通性
		if (row > 1 && grid[row - 2][col - 1] == OPEN) {
			topUF.union((row - 2) * n + col - 1, (row - 1) * n + col - 1);
			bottomUF.union((row - 2) * n + col - 1, (row - 1) * n + col - 1);
		}

		// 与下方节点的连通性
		if (row < n && grid[row][col - 1] == OPEN) {
			topUF.union(row * n + col - 1, (row - 1) * n + col - 1);
			bottomUF.union(row * n + col - 1, (row - 1) * n + col - 1);
		}

		// 与左侧节点的连通性
		if (col > 1 && grid[row - 1][col - 2] == OPEN) {
			topUF.union((row - 1) * n + col - 2, (row - 1) * n + col - 1);
			bottomUF.union((row - 1) * n + col - 2, (row - 1) * n + col - 1);
		}

		// 与右侧节点的连通性
		if (col < n && grid[row - 1][col] == OPEN) {
			topUF.union((row - 1) * n + col, (row - 1) * n + col - 1);
			bottomUF.union((row - 1) * n + col, (row - 1) * n + col - 1);
		}

		/*
		 * 判断条件!percolateFlag是为了防止渗透以后的重复判断 判断条件openedNum>=n
		 * 是因为openedNum达到n时才有可能渗透，在未达到n之前，不需要进行后续判断
		 * 一个节点open的时候刚好使grid渗透的条件是该节点同时与top虚节点和bottom虚节点连通
		 */
		if (!percolateFlag && openedNum >= n && topUF.connected(n * n, (row - 1) * n + col - 1)
				&& bottomUF.connected(n * n, (row - 1) * n + col - 1))
			percolateFlag = true;

	}

	public boolean isOpen(int row, int col) {
		// is site (row, col) open?
		validate(row, col);
		return grid[row - 1][col - 1] == OPEN;
	}

	/**
	 * 一个节点只有同时在open状态并且与top虚节点连通时才是full状态
	 * @param row
	 * @param col
	 * @return
	 */
	public boolean isFull(int row, int col) {
		// is site (row, col) full?
		validate(row, col);
		if (isOpen(row, col) && topUF.connected(n * n, (row - 1) * n + col - 1))
			return true;
		else
			return false;
	}

	public int numberOfOpenSites() {
		// number of open sites
		return openedNum;
	}

	public boolean percolates() {
		// does the system percolate?
		return percolateFlag;
	}

	//打印一些便于查看的信息
	private void printCheckResult(int row, int col) {
		StdOut.println("p(" + row + "," + col + ") is open=" + isOpen(row, col) + ";is full=" + isFull(row, col)
				+ ";percolates=" + percolates());
	}

	/**
	 * 作业提交时main需要调用该方法，因为提交后在线脚本要用一堆input文件进行测试
	 * 
	 * @param arg0
	 */
	private static void fileInputCheck(String arg0) {
		// test client (optional)
		In in = new In(arg0);//读入input文件名，并加载文件内容
		String s = null;
		int n = -1;
		//读入grid的n
		while (in.hasNextLine()) {
			s = in.readLine();
			if (s != null && !s.trim().equals(""))
				break;
		}
		s = s.trim();
		n = Integer.parseInt(s);
		Percolation p = new Percolation(n);
		
		//读入open的site坐标
		while (in.hasNextLine()) {
			s = in.readLine();
			if (s != null && !s.trim().equals("")) {
				s = s.trim();//去掉输入字符串头尾空格
				String[] sa = s.split("\\s+");//去掉中间所有空格
				if (sa.length != 2)
					break;
				int row = Integer.parseInt(sa[0]);
				int col = Integer.parseInt(sa[1]);
				p.open(row, col);
			}
		}

	}

	/**
	 * 本地测试专用
	 */
	private static void generateCheck() {
		// test client (optional)
		Percolation p = new Percolation(3);
		int row = 1, col = 3;
		p.open(row, col);
		p.printCheckResult(row, col);
		row = 2;
		col = 3;
		p.open(row, col);
		p.printCheckResult(row, col);
		row = 3;
		col = 3;
		p.open(row, col);
		p.printCheckResult(row, col);
		row = 3;
		col = 1;
		p.open(row, col);
		p.printCheckResult(row, col);
		row = 2;
		col = 1;
		p.open(row, col);
		p.printCheckResult(row, col);
		row = 1;
		col = 1;
		p.open(row, col);
		p.printCheckResult(row, col);
	}

	public static void main(String[] args) {
		generateCheck();
		// fileInputCheck(args[0]);
	}
}

仿真分析这一部分比较简单，其中需要注意的地方就是“随机选取row和col进行open”，如果简单的用random(int n)，选取[0,n)获取row和col，会有很多重复节点被选中，随着n越大，命中率就越低。于是我采用生成一个[0,n*n)的数组，数组内容随机排序,依次读取数组内容，就相当于随机取site。

empty