【高效算法设计】UVa120 Stack 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 55 4 3 2 15 1 2 3 4

Sample Output

1 2 3 4 505 4 3 2 11 05 1 2 3 4
1 2 0

题意：在一组序列里，给定一种新的排序方式，排序只能操作flip完成，flip(i)表示将数组倒数第i个到1个元素位置颠倒，使用该方式将数组sort为从小到大

思路：flip操作只能对i以前的序列进行操作，也就是说i之后的序列排好序后就可以不再考虑，因为它们已经不受后续flip操作的影响， 并且我们使用该操作可以确保把数组的第一个元素颠倒到位置i去，所以我们每次都把最大的通过flip操作移到最后即可，具体操作是找到最大的数的位置，将其flip到最前面，然后再flip到相应位置即可

由于此题给出的序列值不一定是连续的1~n，所以这里我使用离散化的思想，利用数组MAP，将给定数据映射为1~n

此题的input并没有给出数组个数n的大小，所以只能通过按行读入的方法，这里我采用stringstream+getline的方式进行读取

代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<sstream>#include<algorithm>#include<string>using namespace std;int MAP[1000];int a[1000];int b[1000];string str;void flip(int k){    int temp;    for(int i=1;i<=k/2;i++)    {        temp=a[i];        a[i]=a[k-i+1];        a[k-i+1]=temp;    }}int main(){    int i,j,n,tt;    while(getline(cin,str))    {        stringstream ss(str);        i=1;        while(ss>>a[i])        {            cout<<a[i]<<" ";            b[i]=a[i];            i++;        }        cout<<endl;        n=i-1;        sort(b+1,b+1+n);        for(i=1;i<=n;i++)        {            MAP[b[i]]=i;        }        for(i=n;i>=1;i--)        {            if(MAP[a[i]]==i)                continue;            else            {                for(j=i-1;j>=1;j--)                {                    if(MAP[a[j]]==i)                    {                        if(j!=1)                        {                            cout<<n+1-j<<" ";                            flip(j);                                                  }                        cout<<n+1-i<<" ";                        flip(i);                                     break;                    }                }            }        }        cout<<0<<endl;    }    return 0;}