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 stateprivate 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 blockedif (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blockfor (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 alreadyvalidate(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;elsereturn false;}public int numberOfOpenSites() {// number of open sitesreturn 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nwhile (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