算法(Algorithms)第4版 练习 1.5.1
2024-08-29 22:05:20
id数组的变化情况:
0 1 2 3 4 5 6 7 8 9
10 components
9 0
0 1 2 3 4 5 6 7 8 0
9 components
3 4
0 1 2 4 5 6 7 8 0
8 components
5 8
0 1 2 4 4 6 7 8 0
7 components
7 2
0 1 2 4 4 8 6 8 0
6 components
2 1
0 1 4 4 8 6 8 0
5 components
5 7
0 1 1 4 4 6 1 0
4 components
0 3
1 1 4 4 1 6 1 1
3 components
4 2
1 1 1 6 1 1
2 components
操作次数分析:
find()函数每次调用访问数组1次。
connected函数每次调用两次find()函数,故访问数组2次。
union函数访问数组的次数为:2 + N + (1,N-1)。其中2为两次调用find()函数,N为N次数组判断,(1,N-1)为可能的数组替换次数。
public static void main(String[] args) { //initialize N components
int N = StdIn.readInt();
UFQuickFind uf = new UFQuickFind(N);
StdOut.println(uf); 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");
continue;
} uf.union(p, q);//connect p and q
StdOut.println(p + " " + q);
StdOut.println(uf);
}
对于这个client,对每个数据对,都调用一次connected函数和union函数。
下边对数组访问次数进行分析:
9 0:2 + 2 + 10 + 1
3 4:2 + 2 + 10 + 1
5 8:2 + 2 + 10 + 1
7 2:2 + 2 + 10 + 1
2 1:2 + 2 + 10 + 2
5 7:2 + 2 + 10 + 2
0 3:2 + 2 + 10 + 2
4 2:2 + 2 + 10 + 4
源代码:
package com.qiusongde; import edu.princeton.cs.algs4.StdIn;
import edu.princeton.cs.algs4.StdOut; public class UFQuickFind { private int[] id;//access to component id (site indexed)
private int count;//number of components public UFQuickFind(int n) {
//initialize count and id
count = n; id = new int[n];
for(int i = 0; i < n; i++) {
id[i] = i;
} } public int count() {
return count;
} public boolean connected(int p, int q) {
return find(p) == find(q);
} public int find(int p) {
return id[p];
} public void union(int p, int q) { int pID = find(p);
int qID = find(q); //do nothing if p and q are already
//in the same component
if(pID == qID)
return; //rename p's component to q's name
for(int i = 0; i < id.length; i++) {
if(id[i] == pID)
id[i] = qID;
}
count--; } @Override
public String toString() {
String s = ""; for(int i = 0; i < id.length; i++) {
s += id[i] + " ";
}
s += "\n" + count + " components"; return s;
} public static void main(String[] args) { //initialize N components
int N = StdIn.readInt();
UFQuickFind uf = new UFQuickFind(N);
StdOut.println(uf); 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");
continue;
} uf.union(p, q);//connect p and q
StdOut.println(p + " " + q);
StdOut.println(uf);
} } }
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