贴上详细的题目:
星期五的晚上,一帮微软技术员在希格玛附近的“硬盘酒吧”谈算法问题。有个同事说: 我以前在烙饼店打工,顾客经常端非常多的烙饼。店里的饼大小不一, 我习惯在到达顾客饭桌前,把一摞饼按照大小次序摆好——小的在上面,大的在下面。由于我一只手托着盘子,只好用另一只手,一次抓住最上面的几块饼,把它们 上下颠倒个个儿,反复几次之后,这摞烙饼就排好序了。
假设有n块大小不一的烙饼,那最多/最少要翻几次,才能达到最后大小有序的结果呢? 吧台的酒保说:这太难了吧。我有个简单的问题。有一次我烙了三个饼,一个两面都焦了,一
个两面都是金黄色,一个一面是焦的,一面是金黄色,我把它们摞一起,只能看到最上面一个饼的一面,发现是焦的,问最上面这个饼的另一面是焦的概率是多少? 不少喝酒的人脱口而出:1/2! 上面的说法对吗?你能否写出一个程序,对于n块大小不一的烙饼,输出最优化的翻饼过程呢?
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Code
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usings#region usings
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using System;
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using System.Collections.Generic;
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using System.Linq;
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using System.Text;
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using System.Collections.Generic;
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#endregion
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namespace CakeSorting
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{
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class Program
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{
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Vars#region Vars
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private int[] cakeSizeArray;//大饼直径队列
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private int[] cakeArray;//大饼队列
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private Stack<int> steps;
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private Queue<int> cakes;
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const int cakeNum = 7;
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private int maxTime = 0;
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private int minTime = 0;
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#endregion
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static void Main(string[] args)
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{
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Program p = new Program();
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p.run();
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}
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void run()
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{
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DateTime dt = DateTime.Now;
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TimeSpan ts;
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init();
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enumeratQueue(cakes.Count());
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ts = DateTime.Now - dt;
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Console.WriteLine("\nResult\nMinTime is:{0}\t{1}:{2}:{3}:{4}", minTime, ts.Hours, ts.Minutes, ts.Seconds, ts.Milliseconds);
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Console.ReadLine();
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}
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void init()
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{
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//初始化赋值
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setMaxTime();
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minTime = 0;
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cakeArray = new int[cakeNum];
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cakeSizeArray = new int[cakeNum];
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cakes = new Queue<int>();
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steps = new Stack<int>();
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for (int i = 0; i < cakeSizeArray.Length; i++)
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{
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cakeSizeArray[i] = i;
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}//大饼直径初始化
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foreach (int i in cakeSizeArray)
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{
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cakes.Enqueue(i);
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}//大饼队列化
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}
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void enumeratQueue(int cn)
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{
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if (cn == 1)
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{
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int t = cakes.Dequeue();
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cakeArray[0] = t;
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setMaxTime();
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search(0);
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minTime = maxTime > minTime ? maxTime : minTime;
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show();
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cakes.Enqueue(t);
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}
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for (int i = 0; i < cn; i++)
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{
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int t = cakes.Dequeue();
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cakeArray[cn - 1] = t;
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enumeratQueue(cn - 1);
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cakes.Enqueue(t);
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}
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}
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void search(int step)
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{
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if (maxTime < 15 * cakeNum / 14)
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return;//当此最优解必小于此最劣情况下界时剪枝
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if (step > maxTime)
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return;//大于2(n-1)时退出
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if (step + getLowerTime() > maxTime)
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return;//最优预计
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for (int i = 1; cakeArray [i-1]<cakeArray [i]; i++)
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{
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if (i == cakeArray.Length - 1)
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{
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maxTime = step;
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return;
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}
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}//排序成功退出
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for (int i = 0; i < cakeArray.Length; i++)
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{
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revert(0, i);
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search(step+1);
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revert(0, i);
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}//递归穷举所有方案
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}
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void revert(int begin, int end)
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{
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int t = cakeArray[begin];
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for (int i = begin, j = end; i < j; i++, j--)
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{
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t = cakeArray[i];
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cakeArray[i] = cakeArray[j];
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cakeArray[j] = t;
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}
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}
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void show()
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{
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foreach (int i in cakeArray)
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{
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Console.Write("{0}\t", i);
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}
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Console.Write("\nMaxTime is:{0}\tMinTime is:{1}\n**************************************************\n",maxTime ,minTime);
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}
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void setMaxTime()
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{
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//maxTime = 2 * (cakeNum - 1);
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maxTime = (5 * cakeNum + 5) / 3 < 2 * (cakeNum - 1) ? (5 * cakeNum + 5) / 3 : 2 * (cakeNum - 1);
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}
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int getLowerTime()
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{
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int count=0;
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for (int i = 1; i < cakeNum; i++)
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{
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if (Math.Abs(cakeArray[i - 1] - cakeArray[i]) != 1)
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count++;
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}
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return count;
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}
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}
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}
上一篇写了当已知一个烙饼队列,求出其最少反转次数。
这一篇中将求出如题即n个烙饼最坏情况下需要多少次反转。
《编程之美》中给出目前最优最大下界:15n/14,最小的上界(5n+5)/3。
c#代码:
最优解(已经使用了书中说的当碰到相同状态时根据是否是是更优解的情况判断,程序还可以继续优化,有兴趣的朋友一起讨论):
运行结果: ![](/assets/blank.gif)
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Code
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using System;
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using System.Collections.Generic;
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using System.Linq;
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using System.Text;
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namespace CakeArraySorting
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{
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class Program
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{
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Main#region Main
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static void Main(string[] args)
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{
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Program p = new Program();
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p.run();
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Console.ReadLine();
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}
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#endregion
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Parameters#region Parameters
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private int[] cakeSizeArray;//大饼直径队列
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private int[] cakeArray;//大饼队列
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private int[] maxCakeArray;
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private Queue<int> cakes;
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const int cakeNum = 9;
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private int maxTime = 0;
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private int minTime = 0;
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private int lowwerTime;
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private int ID = 0;
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private int[] timeTable;
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private DateTime startTime;
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private TimeSpan ts;
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#endregion
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Process#region Process
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void run()
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{
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init();
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enumeratQueue(cakes.Count());
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ts = DateTime.Now - startTime;
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Console.Write("MaxCakeArray:\t");
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foreach (int i in maxCakeArray)
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{
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Console.Write("{0}\t", i);
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}
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Console.WriteLine("\nCost Time:{0}\t MaxTime:{1}", ts, minTime);
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}
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void init()
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{
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//初始化赋值
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startTime = DateTime.Now;
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setMaxTime();
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lowwerTime = (int)15 * cakeNum / 14;
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minTime = 0;
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cakeArray = new int[cakeNum];
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cakeSizeArray = new int[cakeNum];
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maxCakeArray = new int[cakeNum];
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cakes = new Queue<int>(cakeNum);
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setCakes();
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cakesEnqueue();
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timeTable = new int[getFactorial(cakeNum)];
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}
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#endregion
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AppendMethods#region AppendMethods
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void enumeratQueue(int cn)
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{
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if (cn == 1)
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{
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int t = cakes.Dequeue();
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cakeArray[0] = t;
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setMaxTime();
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Console.Write("ID:{0}", ID);
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//foreach (int i in cakeArray)
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//{
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// Console.Write("\t{0}", i);
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//}
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if (getNeberNum() == cakeNum - 1)//无相邻直径大小的饼
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{
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setTimeTable();
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search(0);
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minTime = maxTime > minTime ? maxTime : minTime;
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if (minTime == maxTime)
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Array.Copy(cakeArray, maxCakeArray, cakeNum);
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//show();
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}
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else
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{
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Console.WriteLine("jump!
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.");
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}
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ID++;
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cakes.Enqueue(t);
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}
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for (int i = 0; i < cn; i++)
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{
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int t = cakes.Dequeue();
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cakeArray[cn - 1] = t;
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enumeratQueue(cn - 1);
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cakes.Enqueue(t);
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}
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}
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void search(int step)
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{
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//if (cakeNum == 1)
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// return;
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if (step > maxTime)
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return;//大于2(n-1)时退出
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if (step + getNeberNum() > maxTime)
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return;
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int p = getID();
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if (step <= timeTable[p])
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{
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timeTable[p] = step;
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}
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else
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return;
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for (int i = 1; cakeArray[i - 1] < cakeArray[i]; i++)
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{
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if (i == cakeArray.Length - 1)
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{
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maxTime = step;
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return;
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}
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}//排序成功退出
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for (int i = 0; i < cakeArray.Length; i++)
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{
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revert(0, i);
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search(step+1);
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revert(0, i);
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}//递归穷举所有方案
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}
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void revert(int begin, int end)
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{
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int t = cakeArray[begin];
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for (int i = begin, j = end; i < j; i++, j--)
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{
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t = cakeArray[i];
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cakeArray[i] = cakeArray[j];
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cakeArray[j] = t;
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}
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}
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void show()
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{
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Console.Write("\nMaxTime is:{0}\tMinTime is:{1}\n",maxTime ,minTime);
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}
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void setCakes()
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{
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for (int i = 0; i < cakeSizeArray.Length; i++)
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{
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cakeSizeArray[i] = i;
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}//大饼直径初始化,简便期间直接用相邻数字
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}
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void cakesEnqueue()
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{
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foreach (int i in cakeSizeArray)
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{
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cakes.Enqueue(i);
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}//大饼队列化
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}
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void setMaxTime()
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{
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//maxTime = 2 * (cakeNum - 1);
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maxTime = (5 * cakeNum + 5) / 3;
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}
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int getNeberNum()
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{
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int ret = 0;
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for (int i = 1; i < cakeNum;i++ )
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{
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if (Math.Abs(cakeArray[i - 1] - cakeArray[i]) != 1)
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{
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ret++;
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}
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}
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return ret;
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}
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int getMaxLowwerNum()//计算15n/14~=n的最大数,计算得为14
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{
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int i = 1;
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for (; 15 * i / 14 - i < 1; i++) ;
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return i;
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}
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int getFactorial(int n)
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{
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int ret;
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for (ret = n; n > 1; n--)
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ret *= n - 1;
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return ret;
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}
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int getID( )
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{
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int ret = 0;
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for (int i = 0; i < cakeArray.Length; i++)
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{
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ret += getNumSize(i) * getFactorial(cakeNum - i - 1);
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}
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return ret;
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}
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int getNumSize(int p)
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{
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int ret=0;
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for (int i = p + 1; i < cakeArray.Length; i++)
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{
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if (cakeArray[p] < cakeArray[i])
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ret++;
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}
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return ret;
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}
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void setTimeTable()
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{
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for (int i = 0; i < timeTable.Length; i++)
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{
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timeTable[i] = (int)(5 * cakeNum + 5) / 3;
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}
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}
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#endregion
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}
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}