# 程序控

IPPP (Institute of Penniless Peasent-Programmer) Fellow

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Time limit: 3.000 seconds

## Background背景

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问题

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输入

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输出

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.

1 2 3 4 5
5 4 3 2 1
5 1 2 3 4

1 2 3 4 5
0
5 4 3 2 1
1 0
5 1 2 3 4
1 2 0

2 4 1 3 5

4 2 1 3 5

3 1 2 4 5

2 1 3 4 5

## Solution解答

#include <algorithm>
#include <iostream>
#include <iterator>
#include <deque>
#include <string>
#include <sstream>
using namespace std;
//主函数
int main(void) {
//循环处理输入的每组字符串。每次循环一轮要输出最后的0和换行
for (string strLine; getline(cin, strLine); cout << '0' << endl) {
//按要求回应输入的字符串行
cout << strLine << endl;
//构造字符串流，以遍转换为数字
istringstream iss(strLine);
//将字符串转为数字，逆序(最底的在最前)存储在Stack里
deque<int> Stack;
for (int nDiam; iss >> nDiam; Stack.push_front(nDiam));
//从底依次上向进行翻转，保持i上面的都比i小
for (deque<int>::iterator i = Stack.begin(); i != Stack.end(); ++i) {
//找出i上面(包括i)的最大元素
deque<int>::iterator iMax = max_element(i, Stack.end());
//如果最大元素就是i则继续(将i指向上面一个)
if (iMax != i) { //否则要进行需翻转操作
//如果最大的不在最上面，则需先翻转到最上面
if (iMax != Stack.end() - 1) {
reverse(iMax, Stack.end());
//输出翻转的起点
cout << distance(Stack.begin(), iMax) + 1 << ' ';
}
//将最大的从最上面翻转到i的位置上
reverse(i, Stack.end());
//输出翻转的起点
cout << distance(Stack.begin(), i) + 1 << ' ';
}
}
}
return 0;
}


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 作者：王雨濛；新浪微博：@吉祥村码农；来源：《程序控》博客 -- http://www.cnblogs.com/devymex/ 此文章版权归作者所有（有特别声明的除外)，转载必须注明作者及来源。您不能用于商业目的也不能修改原文内容。

posted on 2010-08-15 01:01  Devymex  阅读(6133)  评论(6编辑  收藏  举报