Uva 120 - Stacks of Flapjacks（构造法）

UVA - 120  Stacks of Flapjacks

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Description

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

题解：构造法,选择排序的思想
   把半径的大小与位置对应对应于每叠煎饼数据，
   必须在第一行输出原叠的内容，接下来输出一组翻转动作的序列，使得这一叠煎饼自底向上由大至小的排列，一一对应。输出的每一组翻转动作序列都要由0来结束。一旦一叠煎饼已经排好序，就不能再进行任何翻转

#include <stdio.h>
int a[],b[],n,p,total,Max;
void swap(int x)
{
    int i, j, t;
    for(i=,j=x; j>i; j--,i++)
    {
        t = a[i];
        a[i] = a[j];
        a[j] = t;
    }
}
int main()
{
    while(scanf("%d",&a[])==)
    {
        n = ;
        while()
        {
            if(getchar()!=' ')
                break;
            scanf("%d", &a[n]);
            n++;
        }
        printf("%d",a[]);
        for(int i=; i<n; i++)
            printf(" %d", a[i]);
        printf("\n");
        total=;
        for(int i=; i<n; i++)
        {
            Max=;
            for(int j=n-i-; j>=; j--)
                if(Max<a[j])
                {
                    Max=a[j];
                    p=j;
                }
            if(p!=n-i-)
            {
                if(p!=)
                {
                    swap(p);
                    b[total++]=n-p;
                }
                swap(n-i-);
                b[total++]=i+;
            }
        }
        if(total == )
            printf("%d\n",);
        else
        {
            for(int i=; i<total; i++)
                printf("%d ",b[i]);
            printf("%d\n", );
        }
    }
    return ;
}