HD 5500Reorder the Books



Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 331    Accepted Submission(s): 229

Problem Description

dxy has a collection of a series of books called "The Stories of SDOI",There aren(n≤19) books in this series.Every book has a number from 1 to n.

dxy puts these books in a book stack with the order of their numbers increasing from top to bottom. dxy takes great care of these books and no one is allowed to touch them.

One day Evensgn visited dxy's home, because dxy was dating with his girlfriend, dxy let Evensgn stay at home himself. Evensgn was curious about this series of books.So he took a look at them. He found out there was a story about "Little E&Little Q". While losing himself in the story,he disrupted the order of the books.

Knowing that dxy would be back soon,Evensgn needed to get the books ordered again.But because the books were too heavy.The only thing Evensgn could do was to take out a book from the book stack and and put it at the stack top.

Give you the order of the disordered books.Could you calculate the minimum steps Evensgn would use to reorder the books? If you could solve the problem for him,he will give you a signed book "The Stories of SDOI 9: The Story of Little E" as a gift.

 

Input

There are several testcases.

There is an positive integer T(T≤30) in the first line standing for the number of testcases.

For each testcase, there is an positive integer n in the first line standing for the number of books in this series.

Followed n positive integers separated by space standing for the order of the disordered books,theith integer stands for the ith book's number(from top to bottom).


Hint:
For the first testcase:Moving in the order of book3,book2,book1 ,(4,1,2,3)→(3,4,1,2)→(2,3,4,1)→(1,2,3,4),and this is the best way to reorder the books.
For the second testcase:It's already ordered so there is no operation needed.

 

Output

For each testcase,output one line for an integer standing for the minimum steps Evensgn would use to reorder the books.

 

Sample Input

244 1 2 351 2 3 4 5

 

Sample Output

30

 

Source

BestCoder Round #59 (div.1)

只怪自己太笨了，没领会这一题。

当时想法是：只有最大在第一个或者最后一个的时候不动，其他的都得往前移动，就在这里 想错了。2 3 4 1,根本就不用移动，只需移动1就行

而正确的思路就是：最大的那个数不用动，只需判断（n-1)在不在n的前面，如果在n的前面n-1也不用动，如果不在n的前面n-1肯定得移动到第一位，与此同时不论比n-1小的数在哪，都需从大到小一个一个移动到n-1的前面。

题解：

先可以发现最大的数$n$是不用操作的（其他数操作好了，数"$n$"自然就在最后面了）。 于是我们先找到最大的数"$n$"的位置，从这个位置往前找，直到找到$(n-1)$。假如找到头也没找到$(n-1)$，那么数"$(n-1)$"需要操作，而一旦操作了$(n-1)$，根据前面结论，总共就需要$(n-1)$次操作了;假如找到了(n-1)，那么数"$(n-1)$"也不需要操作（和数"$n$"不需要操作一个道理）。 同理，我们接着从$(n-1)$的位置往前找$(n-2)$,再从$(n-2)$的位置往前找$(n-3)$...假如数$k$找不到了，那么就至少需要$k$次操作。这种做法的复杂度是$O(n)$的

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int t,n;int a[25];int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        int cnt=0;        int l,num=0;        for(int i=0;i<n;i++)            if(a[i]==n)            {                l=i;                break;            }        for(int i=n-1;i>0;i--)        {            int flag=0;            for(int j=l-1;j>=0;j--)            {                if(a[j] == i)                {                    l=j;                    flag=1;                    break;                }            }            if(flag == 0)            {                num=i;                break;            }        }        printf("%d\n",num);        /*        for(int i=n-1;i>0;i--)        {            for(int j=0;j<n;j++)            {                if(a[j]==i)                {                    if(j==(i-1) || j==0)                    {                        continue;                    }                    else                    {                        int temp=a[j];                        for(int k=j;k>0;k--)                            a[k]=a[k-1];                        a[0]=temp;                        cnt++;                    }                }            }        }*/    }    return 0;}