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 commandflip(1).

The 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.

The 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

此题的题意比较难懂（对于我来讲），大意：有一摞煎饼，知道它们的半径，要求通过翻转的方法，使它们从小到大排列（翻转只能向上翻转），输出每次翻转的位置；

解题思路：就像汉诺塔一样，一定要先把最大的放到最下面，然后放次大的。。因此，我们考虑首先找到无序状态中最大的一个，将它翻转到最上面，然后再将它翻转到它应属的位置；

实现手段：运用algorithm中的reverse函数，代码如下（比较拙）：

#include<iostream>
#include<algorithm>
#include<sstream>
using namespace std;


int nu[50],no[50],n;
string str;


bool bmp(int a,int b){
    return b<a;
}


bool cmp(){
   //sort(nu,nu+n);
    for(int i=0;i<n;i++){
       if(nu[n-i-1]!=no[i])
        return false;
    }
    return true;
}


void solve(){
    int num=n,j;
    for(int i=0;i<n;i++){
        for(j=0;j<n;j++){
            if(no[j]==nu[i]){
                //cout<<j<<endl;
                break;
            }
        }
        if(j==num-1){//当j在它应属的位置时，不需要移动，不一定是最后位置：j=n-1false;
            num--;
        }
        else if(j==0){
            reverse(no,no+num);
            //for(int i=0;i<n;i++) cout<<no[i]<<" ";
            cout<<n-num+1<<" ";
            num--;
        }
        else{
            cout<<n-j<<" ";//翻转的起始位置在最后一个元素而不是第一个元素
            reverse(no,no+j+1);//注意后面加的是翻转元素的数量而不是下标
            reverse(no,no+num);
            cout<<n-num+1<<" ";
            num--;//只要将一个元素归位后，则接下来可翻转的元素的个数就减一
        }
        if(cmp()) break;
    }
    cout<<0<<endl;
}


int main(){
    while(getline(cin,str)){
        stringstream ss;
        int a,i=0,j=0;
        ss<<str;
        while(ss>>a){
           nu[i++]=a;
           no[j++]=a;
        }
        n=i;
        sort(nu,nu+n,bmp);
        for(i=0;i<n;i++){
            if(i) cout<<" ";
            cout<<no[i];
        }
        cout<<endl;
        solve();


    }
    return 0;
}

总结：类似题目要先找到排序规律，可以先从简单的入手。