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算法(Algorithms)第4版 练习 1.5.4

代码实现:

package com.qiusongde;

import edu.princeton.cs.algs4.StdIn;
import edu.princeton.cs.algs4.StdOut;

public class UFWeightedQuickUnion {

    private int[] id;//parent link(site indexed)
    private int[] treesize;//size of component for roots(site indexed)
    private int count;//number of components
    
    private static int totalcost = 0;//for the test of array access
    
    public UFWeightedQuickUnion(int N) {
        
        count = N;
        
        id = new int[N];
        for(int i = 0; i < N; i++) 
            id[i] = i;
        
        treesize = new int[N];
        for(int i = 0; i < N; i++)
            treesize[i] = 1;
        
    }
    
    public int count() {
        return count;
    }
    
    public boolean connected(int p, int q) {
        return find(p) == find(q);
    }
    
    public int find(int p) {
        
        while(p != id[p]) {
            p = id[p];
            totalcost += 2;
        }
        totalcost++;
            
        return p;
        
    }
    
    public void union(int p, int q) {
        
        int pRoot = find(p);
        int qRoot = find(q);
        
        if(pRoot == qRoot)
            return;
        
        //make smaller root point to larger one
        if(treesize[pRoot] < treesize[qRoot]) {
            id[pRoot] = qRoot;
            treesize[qRoot] += treesize[pRoot];
            totalcost += 5;
        } else {
            id[qRoot] = pRoot;
            treesize[pRoot] += treesize[qRoot]; 
            totalcost += 5;
        }
        
        count--;
        
    }
    
    @Override
    public String toString() {
        String s = "";
        
        for(int i = 0; i < id.length; i++) {
            s += id[i] + " ";
        }
        s += "\n";
        
        for(int i = 0; i < treesize.length; i++) {
            s += treesize[i] + " ";
        }
        s += "\n" + count + " components";
        
        return s;
    }
    
    public static void main(String[] args) {
        
        //initialize N components
        int N = StdIn.readInt();
        UFWeightedQuickUnion uf = new UFWeightedQuickUnion(N);
        StdOut.println(uf);
        StdOut.println();
        
        while(!StdIn.isEmpty()) {
            
            int p = StdIn.readInt();
            int q = StdIn.readInt();
            
            if(uf.connected(p, q)) {//ignore if connected
                StdOut.println(p + " " + q + " is connected");
                StdOut.println("totalcost: " + totalcost);
                StdOut.println();
                continue;
            }
            
            uf.union(p, q);//connect p and q
            StdOut.println(p + " " + q);
            StdOut.println(uf);
            StdOut.println("totalcost: " + totalcost);
            StdOut.println();
        }
        
    }
    
}

 

reference input:

10
4 3
3 8
6 5
9 4
2 1
8 9
5 0
7 2
6 1
1 0
6 7

 

结果:

0 1 2 3 4 5 6 7 8 9 
1 1 1 1 1 1 1 1 1 1 
10 components

4 3
0 1 2 4 4 5 6 7 8 9 
1 1 1 1 2 1 1 1 1 1 
9 components
totalcost: 9

3 8
0 1 2 4 4 5 6 7 4 9 
1 1 1 1 3 1 1 1 1 1 
8 components
totalcost: 22

6 5
0 1 2 4 4 6 6 7 4 9 
1 1 1 1 3 1 2 1 1 1 
7 components
totalcost: 31

9 4
0 1 2 4 4 6 6 7 4 4 
1 1 1 1 4 1 2 1 1 1 
6 components
totalcost: 40

2 1
0 2 2 4 4 6 6 7 4 4 
1 1 2 1 4 1 2 1 1 1 
5 components
totalcost: 49

8 9 is connected
totalcost: 55

5 0
6 2 2 4 4 6 6 7 4 4 
1 1 2 1 4 1 3 1 1 1 
4 components
totalcost: 68

7 2
6 2 2 4 4 6 6 2 4 4 
1 1 3 1 4 1 3 1 1 1 
3 components
totalcost: 77

6 1
6 2 6 4 4 6 6 2 4 4 
1 1 3 1 4 1 6 1 1 1 
2 components
totalcost: 90

1 0 is connected
totalcost: 98

6 7 is connected
totalcost: 104

 

worst-case input:

8
0 1
2 3
4 5
6 7
0 2
4 6
0 4

 

结果:

0 1 2 3 4 5 6 7 
1 1 1 1 1 1 1 1 
8 components

0 1
0 0 2 3 4 5 6 7 
2 1 1 1 1 1 1 1 
7 components
totalcost: 9

2 3
0 0 2 2 4 5 6 7 
2 1 2 1 1 1 1 1 
6 components
totalcost: 18

4 5
0 0 2 2 4 4 6 7 
2 1 2 1 2 1 1 1 
5 components
totalcost: 27

6 7
0 0 2 2 4 4 6 6 
2 1 2 1 2 1 2 1 
4 components
totalcost: 36

0 2
0 0 0 2 4 4 6 6 
4 1 2 1 2 1 2 1 
3 components
totalcost: 45

4 6
0 0 0 2 4 4 4 6 
4 1 2 1 4 1 2 1 
2 components
totalcost: 54

0 4
0 0 0 2 0 4 4 6 
8 1 2 1 4 1 2 1 
1 components
totalcost: 63

 

算法(Algorithms)第4版 练习 1.5.4