Coursera Algorithms Programming Assignment 1: Percolation（100分）

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。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;
 /**
  * @author evasean www.cnblogs.com/evasean/
  */
 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。

 import edu.princeton.cs.algs4.StdOut;
 import edu.princeton.cs.algs4.StdRandom;
 import edu.princeton.cs.algs4.StdStats;
 /**
  * @author evasean www.cnblogs.com/evasean/
  */
 public class PercolationStats {
     /* t fractions 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 int t;//尝试次数
     private final double[] fractions;//每一次尝试的渗透率得分

     private double mean;
     private double stddev;

     public PercolationStats(int n, int trials) {
         // perform trials independent experiments on an n-by-n grid
         if (n <= 0 || trials <= 0)
             throw new IllegalArgumentException("n ≤ 0 or trials ≤ 0");
         t = trials;
         fractions = new double[t];
         for (int i = 0; i < t; i++) {//t次尝试
             Percolation p = new Percolation(n);
             int openNum = 0;
              //为了实现随机open一个site，模仿QuickUnion的定位方法
             //先生成一个[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
             int[] rand = StdRandom.permutation(n * n);
             for (int pos : rand) {
                 //pos = (row-1)*n + col -1
                 int row = pos / n + 1;
                 int col = pos % n + 1;
                 p.open(row, col);
                 openNum++;
                 //只有openNum>=n时才有判断是否渗透的必要
                 if (openNum >= n && p.percolates())
                     break;
             }
             double pt = (double) openNum / (n * n);//单次尝试的渗透率
             fractions[i] = pt;
         }
         /* 作业提交时的某个测试案例要求mean()、stddev()、confidenceLo()、confidenceHi()
          * 在任何时候任何次序调用的情况下都必须返回相同的值，故需要在构造函数中计算mean和stddev
          */
         //作业提交时的某个测试案例要调用一次StdStats.mean方法
         mean = StdStats.mean(fractions);
         //作业提交时的某个测试案例要求要调用一次StdStats.stddev方法
         stddev = StdStats.stddev(fractions);
     }

     public double mean() {
         // sample mean of percolation threshold
         return mean;
     }

     public double stddev() {
         // sample standard deviation of percolation threshold
         return stddev;
     }

     public double confidenceLo() {
         // low endpoint of 95% confidence interval
         return mean - 1.96 * stddev / Math.sqrt(t);
     }

     public double confidenceHi() {
         // high endpoint of 95% confidence interval
         return mean + 1.96 * stddev / Math.sqrt(t);
     }

     public static void main(String[] args) {
         // test client (described below)
         int n = Integer.parseInt(args[0]);
         int t = Integer.parseInt(args[1]);
         PercolationStats ps = new PercolationStats(n, t);
         StdOut.printf("%-25s %s %f \n", "means", "=", ps.mean());
         StdOut.printf("%-25s %s %f \n", "stddev", "=", ps.stddev());
         StdOut.printf("%-25s %s%f%s%f%s\n", "95% confidence interval", "= [", ps.confidenceLo(), ", ",
                 ps.confidenceHi(), "]");
     }
 }