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).

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

#include<cstdio>
#include<iostream>
#include<sstream>
#include<algorithm>
using namespace std; const int maxn = 30 + 5;
int n, a[maxn]; void flip(int p)
{
for(int i = 0; i < p-i; i++)
swap(a[i], a[p-i]);
printf("%d ", n-p);
} int main()
{
string s;
while(getline(cin, s))
{
cout << s << "\n";
stringstream ss(s);
n = 0;
while(ss >> a[n]) n++;
cout<<n<<endl;
for(int i = n-1; i > 0; i--)
{
//找到当前最大值所在的位置;
int p = max_element(a, a+i+1) - a;
cout<<p<<endl;
//假设最大值与它所在的位置同样,继续下一个;
if(p == i) continue;
//转换;将最大值放到第一个位置;
if(p > 0) flip(p);
flip(i);
}
printf("0\n");
}
return 0;
}

Stacks of Flapjacks的更多相关文章

  1. uva 120 stacks of flapjacks ——yhx

     Stacks of Flapjacks  Background Stacks and Queues are often considered the bread and butter of data ...

  2. UVaOJ 120 - Stacks of Flapjacks

    120 - Stacks of Flapjacks 题目看了半天......英语啊!!! 好久没做题...循环输入数字都搞了半天...罪过啊!!! 还是C方便一点...其实C++应该更方便的...C+ ...

  3. Uva 120 - Stacks of Flapjacks(构造法)

    UVA - 120  Stacks of Flapjacks Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld &a ...

  4. uva Stacks of Flapjacks

                                                     Stacks of Flapjacks  题目链接:Click Here~ 题目描写叙述:     ...

  5. 【思维】Stacks of Flapjacks

    [UVa120] Stacks of Flapjacks 算法入门经典第8章8-1 (P236) 题目大意:有一个序列,可以翻转[1,k],构造一种方案使得序列升序排列. 试题分析:从插入排序即可找到 ...

  6. Stacks of Flapjacks(栈)

     Stacks of Flapjacks  Background Stacks and Queues are often considered the bread and butter of data ...

  7. UVa120 - Stacks of Flapjacks

    Time limit: 3.000 seconds限时:3.000秒 Background背景 Stacks and Queues are often considered the bread and ...

  8. B - Stacks of Flapjacks UVA - 120

    BackgroundStacks and Queues are often considered the bread and butter of data structures and find us ...

  9. UVa 120 Stacks of Flapjacks【构造法】

    题意:给出n张煎饼,从上到下输入,每张煎饼上面都有一个数字,厨师每次可以选择第k张煎饼,进行翻转操作,设计一种方法使得所有煎饼按照从小到大排序(最上面的煎饼最小) 首先是这个翻转的操作,如下图 如图所 ...

随机推荐

  1. 洛谷P2234 [HNOI2002]营业额统计(01Tire树)

    题目描述 Tiger最近被公司升任为营业部经理,他上任后接受公司交给的第一项任务便是统计并分析公司成立以来的营业情况. Tiger拿出了公司的账本,账本上记录了公司成立以来每天的营业额.分析营业情况是 ...

  2. Maven用法

    Maven 的使用教程   1.什么是 Maven? Maven项目对象模型(POM),可以通过一小段描述信息来管理项目的构建,报告和文档的软件项目管理工具.   2.下载 Maven? ①.官网下载 ...

  3. SGU 253 Theodore Roosevelt 快速判断点是否在凸包内

    http://acm.sgu.ru/problem.php?contest=0&problem=253 题意简单易懂...给你n个点的凸包(经测试已经是极角序)...判断m个点是否在凸包内.. ...

  4. Office2010激活工具

    mini-KMS Activator是一款好用Office2010激活工具,“KMS”是微软对于“大客户”(VOL或VL)提供的Microsoft产品激活方式之一.在适用此方式的Microsoft产品 ...

  5. [Angular] Custom directive Form validator

    Create a directive to check no special characters allowed: import {Directive, forwardRef} from '@ang ...

  6. 内网使用 IPV6 之 TunnelBroker隧道(6in4)篇

    内网使用 IPV6 之 TunnelBroker隧道(6in4)篇 据非专业网民推测 tunnelbroker isatap 和 6to4  貌似都需要公网,但有网民测试这位大作的方法可行.特别之处是 ...

  7. touch、touchevent-事件总结

    1.废话不多说,直接上代码,如下 package com.example.mygestrue; import android.support.v7.app.ActionBarActivity; imp ...

  8. 给已有数据的oracle表建立外键关系

    PS:这里是给自己做个备忘,下次遇到同类问题的时候,方便查找: 客户在有主外键关系的2张表进行页面删除时报错已有子记录,运维后台处理的时候应该找出相应的数据,先删除子记录,在删主表记录:但客户要的急, ...

  9. 洛谷 P1601 A+B Problem(高精)

    P1601 A+B Problem(高精) 题目背景 无 题目描述 高精度加法,x相当于a+b problem,[b][color=red]不用考虑负数[/color][/b] 输入输出格式 输入格式 ...

  10. 中英文对照 —— 标点符号(punctuation)

    有限的几个: What Are the Fourteen Punctuation Marks in English Grammar? period:句号:comma:逗号:冒号:colon:分号:se ...