【Jason's_ACM_解题报告】Stacks of Flapjacks

Stacks of Flapjacks 

Background
Stacks and Queues are often considered the bread and butter of data structures and find use in architecture, parsing, operating systems, and discrete event simulation. Stacks are also important in the theory of formal languages.


This problem involves both butter and sustenance in the form of pancakes rather than bread in addition to a finicky server who flips pancakes according to a unique, but complete set of rules.


The Problem
Given a stack of pancakes, you are to write a program that indicates how the stack can be sorted so that the largest pancake is on the bottom and the smallest pancake is on the top. The size of a pancake is given by the pancake's diameter. All pancakes in a stack have different diameters.


Sorting a stack is done by a sequence of pancake ``flips''. A flip consists of inserting a spatula between two pancakes in a stack and flipping (reversing) the pancakes on the spatula (reversing the sub-stack). A flip is specified by giving the position of the pancake on the bottom of the sub-stack to be flipped (relative to the whole stack). The pancake on the bottom of the whole stack has position 1 and the pancake on the top of a stack of n pancakes has position n.


A stack is specified by giving the diameter of each pancake in the stack in the order in which the pancakes appear.


For example, consider the three stacks of pancakes below (in which pancake 8 is the top-most pancake of the left stack):


         8           7           2
         4           6           5
         6           4           8
         7           8           4
         5           5           6
         2           2           7
The stack on the left can be transformed to the stack in the middle via flip(3). The middle stack can be transformed into the right stack via the command flip(1).


Input
The input consists of a sequence of stacks of pancakes. Each stack will consist of between 1 and 30 pancakes and each pancake will have an integer diameter between 1 and 100. The input is terminated by end-of-file. Each stack is given as a single line of input with the top pancake on a stack appearing first on a line, the bottom pancake appearing last, and all pancakes separated by a space.


Output
For each stack of pancakes, the output should echo the original stack on one line, followed by some sequence of flips that results in the stack of pancakes being sorted so that the largest diameter pancake is on the bottom and the smallest on top. For each stack the sequence of flips should be terminated by a 0 (indicating no more flips necessary). Once a stack is sorted, no more flips should be made.


Sample Input
1 2 3 4 5
5 4 3 2 1
5 1 2 3 4


Sample Output
1 2 3 4 5
0
5 4 3 2 1
1 0
5 1 2 3 4
1 2 0

这些天做难题做多了导致思维僵化，一直以为是否有什么高效的方法可以不通过反转来确定最终的答案；而且没有看清题意，以为题目要求给出最优解，其实，根本就没有所谓的最优解，最优解就是模拟题目的描述。

另外Liu及网友们都说这是选择排序的思想，其实可以抛开多余的想法，仅仅看数据规模是30那么我们可以放心的对其进行模拟，当然了，最后做完题在结合对选择排序的理解，选择排序是交换排序中最好的一种，他仅仅需要最多进行n-1次交换，我们发现选择排序的思想是很有用的。

附代码如下：

#include<string>#include<sstream>#include<iostream>#include<algorithm>using namespace std;#define MAXN (30+5)string str;int n,a[MAXN];bool input(){if(!getline(cin,str)) return false;cout<<str<<endl;stringstream ss(str);n=0;while(ss>>a[n])n++;return true;}void flip(int x){for(int i=0;i<=x/2;i++)swap(a[i],a[x-i]);cout<<n-x<<' ';}int main(){while(input()){for(int i=n-1;i>0;i--){int p=max_element(a,a+i+1)-a;if(p==i)continue;if(p>0)flip(p);flip(i);}cout<<0<<endl;}return 0;}