132.1.001 Union-Find | 并查集
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@(132 - ACM | 算法)
Algorithm | Coursera - by Robert Sedgewick
> Tip: Focus on WHAT is really important!
> Don't just copy it!
> Don't look at the subtitle
> Practice is the key. Just Do it!
Backup
P.S. iff == if and only if
0 Introduction
- Dynamic connectivity problem
- union-find data type
- quick find
- quick union
- improvements
- weighted quick union
- weighted quick union with path compression
- applications
- Maze Problem
1 Steps to developing a usable algorithm
- Model the problem
- Find a algorithm
- Fast enough? Fits in memory?
- if not, figure out why
- Find a way to address the problem
- Iterate until satisfied
2 Quick Find | 快速查找
Structure - Linear
Java implementation
public class QuickFindUF
{
private int[] id;
//constructor
public QuickFindUF(int N)
{
id = new int[N];//allocate N*int
for (int i = 0;i<N;i++)
id[i] = i;
}
public boolean connected(int p,int q)
{
return id[p] == id[q];
}
public void union(int p, int q)
{
int pid = id[p];
int qid = id[q];
for (int i = 0;i<id.length;i++)
{
if(id[i]==pid) id[i] = qid;
}
}
}
Quick find is too slow
3 Quick Union
Structure-Tree
- Check if they have the same root
inspire: use the third standard as Reference//第三方标准作为参照物,语言同理
Java implementation
public class QuickUnionUF
{
private int[] id;
//constructor —— set each element to be its own root
public QuickUnionUF(int N)
{
id = new int[N];
for (int i = 0;i < N;i++) id[i] = i;
}
//find the root by chasing parent pointers
private int root(int i)
{
while (i != id[i] i = id[i]);
return i;
}
public boolean connected(int p, int q)
{
return root(p) == root(q);
}
public void union(int p,int q)
{
int i = root(p);
int j = root(q);
id[i] = j;
}
}
Quick Union is also too slow
4 Quik-Union Improvement1 -Weighted quick-union
smaller tree down below - small depends on the bigger( size)
demo
improvement
Java implementation
//Data Structure
//maintain extra array sz[i] to count number of objects in the tree rooted at i
//sz[i] = size of tree rooted i
// Find - identical to quick-union
//Union
//Modify quick-union to:
//1.Link root of smaller tree to root of larger tree.
//2.Update the sz[] array
int i = root(p);
int j = root(q);
if (i == j) return;
if (sz[i] < sz[j]) { id[i] = j; sz[j] += sz[i]};
else { id[j] = i; sz[i] += sz[j]};
Runing time
O(N) = lg N
5 Quik-Union Improvement2 -Path compression
Flatten the tree
In practice - Keeps tree almost completely flat.
java implementation
- Make every other node in path point to its grandparent
private int root(int i)
{
while(i != id[i])
{
id[i] = id[id[i]];//only one extra line of code!
i = id[i];
}
return i;
}
O(N) = N+M * lgN
6 Summary for solving the dynamic connectivity problem
7 Union-Find Application
Percolation
Monte Carlo simulation //蒙特卡罗模拟
Dynamic connectivity solution to estimate percolation threshold
- Clever Trick
8 Application - Percolation | 渗滤问题
需要注意的是Timing和Backwash的问题
Timing:PercolationStats.java里StdRandom.mean()和StdRandom.stddev()都只能调用一次;Percolation.java里实现numberOfOpenSites时切记不能使用循环累加,定义一个私有属性来计数即可;实现open()时相邻的四个sites位置直接加减n或1即可。
Backwash:实现isFull()时需要另外实例化一个不包含最下端虚拟节点的WeightedQuickUnionUF,可以解决Test 13: check for backwash with predetermined sites,Test 14: check for backwash with predetermined sites that have multiple percolating paths和Test 15: call all methods in random order until all sites are open, allowing isOpen() to be called on a site more than once
三项测试无法通过的问题。Backwash问题是指因为虚拟底部结点的存在,导致底部任一结点渗漏成功的话底部所有结点都会认为渗漏成功。原因是通过底部虚拟结点形成了回流。从而导致isFull()方法出错。
- Percolation.java
import edu.princeton.cs.algs4.WeightedQuickUnionUF;
public class Percolation {
private boolean[] op; // true=open while false=blocked
private int side; // number of rows or columns
private int numOp; // number of open sites
private WeightedQuickUnionUF uf;
private WeightedQuickUnionUF ufTop;
public Percolation(int n) {
if(n <= 0) throw new IllegalArgumentException("Input should be positif!
");
this.side = n;
this.op = new boolean[n*n+2]; // with 2 virtual sites
this.uf = new WeightedQuickUnionUF(n*n+2);
this.ufTop = new WeightedQuickUnionUF(n*n+1); // with only the upper virtual site
for(int i=1; i<n*n+1; i++) op[i] = false;
op[0] = op[n*n+1] = true;
this.numOp = 0;
}
// both ROW and COL should be integer within 1~n
private void checkBounds(int row, int col){
if(row < 1 || row > this.side || col < 1 || col > this.side){
throw new IllegalArgumentException("Index out of bounds!
");
}
}
// get position of sites in 3 arrays: op, uf.parent & uf.size
private int getPosition(int row, int col){
return (row - 1) * this.side + col;
}
private void union(int aPos, int bPos, WeightedQuickUnionUF wq){
if(!wq.connected(aPos, bPos)){
wq.union(aPos, bPos);
}
}
private boolean isOpen(int pos){
return op[pos];
}
public void open(int row, int col) {
checkBounds(row, col);
if(isOpen(row, col)) return;
int pos = getPosition(row, col);
op[pos] = true;
numOp++;
// positions of adjacent sites
int rowPrev = pos - side, rowNext = pos + side,
colPrev = pos - 1, colNext = pos + 1;
// try connect the adjacent open sites
if(row == 1){
union(0, pos, uf);
union(0, pos, ufTop);
}else if(isOpen(rowPrev)){
union(rowPrev, pos, uf);
union(rowPrev, pos, ufTop);
}
if(row == side){
union(side * side + 1, pos, uf);
}else if(isOpen(rowNext)){
union(rowNext, pos, uf);
union(rowNext, pos, ufTop);
}
if(col != 1 && isOpen(colPrev)) {
union(colPrev, pos, uf);
union(colPrev, pos, ufTop);
}
if(col != side && isOpen(colNext)) {
union(colNext, pos, uf);
union(colNext, pos, ufTop);
}
}
public boolean isOpen(int row, int col) {
checkBounds(row, col);
return isOpen(getPosition(row, col));
}
/**
* check for backwash with predetermined sites that have multiple percolating paths
* in this case ufTop should be used instead of uf
* @param row
* @param col
* @return
*/
public boolean isFull(int row, int col) {
checkBounds(row, col);
//return uf.connected(0, getPosition(row, col)); -> didn't pass the test!
return ufTop.connected(0, getPosition(row, col));
}
// should pass the timing check
public int numberOfOpenSites(){
return this.numOp;
}
public boolean percolates(){
return uf.connected(0, side * side + 1);
}
}
- PercolationStats.java
import edu.princeton.cs.algs4.StdIn;
import edu.princeton.cs.algs4.StdOut;
import edu.princeton.cs.algs4.StdRandom;
import edu.princeton.cs.algs4.StdStats;
import edu.princeton.cs.algs4.Stopwatch;
public class PercolationStats {
private double[] results; // estimated threshold for each trial
private double avg;
private double std;
public PercolationStats(int n, int trials){
if(n <= 0 || trials <= 0) throw new IllegalArgumentException();
results = new double[trials];
for(int i = 0; i < trials; i++){
int step = 0;
Percolation pr = new Percolation(n);
while(!pr.percolates()){
int row = StdRandom.uniform(n) + 1;
int col = StdRandom.uniform(n) + 1;
if(!pr.isOpen(row, col)){
pr.open(row, col);
step++;
}
}
results[i] = (double)step / (n * n);
}
this.avg = StdStats.mean(results);
this.std = StdStats.stddev(results);
}
public static void main(String[] args){
StdOut.printf("%-25s
", "Please input 2 integers");
int N = StdIn.readInt();
int T = StdIn.readInt();
Stopwatch wt = new Stopwatch();
PercolationStats ps = new PercolationStats(N, T);
// elapsed CPU time in seconds
double elapsed = wt.elapsedTime();
StdOut.printf("%-25s= %.15f
", "elapsed CPU time", elapsed);
StdOut.printf("%-25s= %.7f
", "mean", ps.mean());
StdOut.printf("%-25s= %.17f
", "stddev", ps.stddev());
StdOut.printf("%-25s= [%.15f, %.15f]
", "%95 confidence interval",
ps.confidenceLo(), ps.confidenceHi());
}
public double mean(){
return this.avg;
}
public double stddev(){
return this.std;
}
public double confidenceLo(){
return mean() - 1.96 * stddev() / Math.sqrt(results.length);
}
public double confidenceHi(){
return mean() + 1.96 * stddev() / Math.sqrt(results.length);
}
}
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