UVA120 - Stacks of Flapjacks

Stacks of Flapjacks

Time limit: 3.000 seconds

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
         7           8           4
         6           4           8
         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
1 0
5 4 3 2 1
1 2 0
5 1 2 3 4
题意:  锅子里面有一叠煎饼,一共n张,每张煎饼标有一个数字,厨师可以选择一个数k,把从锅底数第k张上面的煎饼全部翻过来,即原来在上面的煎饼被翻到了下面！算法分析：    题目要求是对煎饼序号排序,实际是"颠倒一个连续子序列",题目可以用到选择排序思想，以从大到小的顺序依次把每个数排到正确的位置,即每次找出前 n-i-1 个数中最大的一个数(0<=i<n)，将最大数以及最大数前面的进行一次翻转，使得最大数在最上面,然后再将 n-i-1 之前所有煎饼翻转一次,使得当前最大数排到正确位置！代码实现：
#include <stdio.h>#include <iostream>#include <algorithm>#include <math.h>#include <string.h>using namespace std;int a[1000];void Swap(int z)///将0-z的位置的值进行翻转{    int temp;    for(int i = 0, j = z; i < j; i++, j--)    {        temp = a[i];        a[i] = a[j];        a[j] = temp;    }}int main(){    int ans[1000];    while(~scanf("%d",&a[0]))    {        int n = 1, num = 0;        while(1)        {            if(getchar() != ' ')                break;            scanf("%d",&a[n++]);        }        printf("%d",a[0]);        for(int i = 1; i < n; i++)            printf(" %d",a[i]);        printf("\n");        int flag = 0;        for(int i = 0; i < n; i++)        {            int Max = -10000;            for(int j = n - i - 1; j >= 0; j--)            {                if(a[j] > Max)///在n-i-1之前找到一个最大数                {                    Max = a[j];                    flag = j;                }            }            if(flag != n - i -1)///最大值不在末尾            {                if(flag != 0)///最大值不在开头                {                    Swap(flag);                    ans[num++] = n - flag;///记录下翻煎饼的位置                }                Swap(n - i - 1);                ans[num++] = i + 1;            }        }        if(num == 0)            printf("0\n");        else        {            for(int i = 0; i < num; i++)                printf("%d ",ans[i]);            printf("0\n");        }    }    return 0;}