一摞烙饼

题目描述

有一摞烙饼，大小不一，编号大小表示了其大小。现在我们需要把这一摞烙饼进行排序，使得从上到下越来越大。但是排序能够进行的操作只有抓起一摞饼进行上下翻转。比如一摞饼是[1,3,5,2,0]，需要的结果是[0,1,2,3,5]。能够进行比如[5,3,1,2,0]，然后再翻转一次就是[0,2,1,3,5]。这种情况下，最多需要2(n-1)次翻转。

代码实现

#include "stdafx.h"#include <iostream>#include <cassert>using namespace std;class CPrefixSorting {public:    CPrefixSorting() : cakesCount(0), cakesArray(NULL), searchCount(0),       curCakesArrayReverse(NULL), cakesArrayReverse(NULL), reversesCount(0) {}    ~CPrefixSorting();    void Run(int *, int);private:    void Init(int *, int);    void Search(int *, int);    void PrintResult(void);    int UpperBound();    int LowerBound(int *);    bool IsSorted(int *);    void Reverse(int *, int, int);private:    int cakesCount;          // The number of cakes.    int *cakesArray;     // The cakes array.    int searchCount;     // The counter of search steps.    int *curCakesArrayReverse;  // The current reverse information.    int *cakesArrayReverse;  // The result reverse information.    int reversesCount;          // The counter of reverses.};CPrefixSorting::~CPrefixSorting() {    if (cakesArray)  delete [] cakesArray;    if (curCakesArrayReverse)  delete [] curCakesArrayReverse;    if (cakesArrayReverse)     delete [] cakesArrayReverse;}void CPrefixSorting::Init(int *pCakes, int cnt) {    assert(NULL != pCakes && cnt > 0);    cakesCount = cnt;    cakesArray = new int[cakesCount];    assert(NULL != cakesArray);    for (int i = 0; i < cakesCount; ++i) {       cakesArray[i] = pCakes[i];    }    reversesCount = UpperBound();    curCakesArrayReverse = new int[reversesCount];    assert(NULL != curCakesArrayReverse);    cakesArrayReverse = new int[reversesCount];    assert(NULL != cakesArrayReverse);    searchCount = 0;}int CPrefixSorting::UpperBound() {    return 2 * cakesCount;}int CPrefixSorting::LowerBound(int *pCakesArray) {    int count = 0;    for (int i = 0; i < cakesCount - 1; ++i)    {       int diff = pCakesArray[i] - pCakesArray[i + 1];       if (-1 != diff && 1 != diff)       {           ++count;       }    }    return count;}bool CPrefixSorting::IsSorted(int *pCakesArray){    for (int i = 0; i < cakesCount - 1; ++i)    {       if (pCakesArray[i] > pCakesArray[i + 1])        {           return false;       }    }    return true;}void CPrefixSorting::Reverse(int *pCakesArray, int start, int end){    assert(start < end);    while (start < end)    {       int tmp = pCakesArray[start];       pCakesArray[start] = pCakesArray[end];       pCakesArray[end] = tmp;       ++start;       --end;    }}void CPrefixSorting::Search(int *pCakesArray, int step){    ++searchCount;    if (LowerBound(pCakesArray) + step >= reversesCount)    {       return;    }    if (IsSorted(pCakesArray) && step < reversesCount)    {       reversesCount = step;       for (int i = 0; i < reversesCount; ++i)       {           cakesArrayReverse[i] = curCakesArrayReverse[i];           cout << curCakesArrayReverse[i] << "  ";       }       cout << endl;       return;    }    for (int i = 1; i < cakesCount; ++i)    {       int *cakesArrayTmp = new int[cakesCount];       assert(NULL != cakesArrayTmp);       for (int j = 0; j < cakesCount; ++j)       {           cakesArrayTmp[j] = pCakesArray[j];       }       Reverse(cakesArrayTmp, 0, i);       curCakesArrayReverse[step] = i;       Search(cakesArrayTmp, step + 1);       if (cakesArrayTmp)        {           delete [] cakesArrayTmp;       }    }}void CPrefixSorting::PrintResult(){    cout << "The search times is " << searchCount << endl;    cout << "The least reverse times is " << reversesCount << endl;    cout << "The reverse process is:" << endl;    for (int i = 0; i < reversesCount; ++i)    {       cout << cakesArrayReverse[i] << "  ";    }    cout << endl;}void CPrefixSorting::Run(int *pCakes, int cnt){    // 初始化     Init(pCakes, cnt);    // 搜索    Search(cakesArray, 0);    // 打印结果    PrintResult();}int _tmain(int argc, _TCHAR* argv[]){    // 需要排列的数组    int arry[] = {6, 8, 9, 5, 1, 7, 11, 10, 2, 3, 4, 12};    CPrefixSorting prefixSorting;    prefixSorting.Run(arry, sizeof(arry) / sizeof(arry[0]));    return 0;}

对于三个饼，一个全焦、一个半焦、一个完好。第一面是焦，那么是全焦的概率为： 2/3。

参考链接：

http://blog.csdn.net/yahohi/article/details/7448676