hdu 1394 Minimum Inversion Number 归并排序求逆序数求一串数的最小逆序数

最新推荐文章于 2019-02-25 23:17:34 发布

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最新推荐文章于 2019-02-25 23:17:34 发布

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分类专栏： acm_数学问题 acm_数据结构文章标签： permutation merge numbers output integer input

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本文介绍了如何利用归并排序算法计算一串整数的最小逆序数。在时间限制2000ms和内存限制65536K的条件下，解决HDU 1394问题，共有927次提交，其中514次成功接受了答案。

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Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 927 Accepted Submission(s): 514

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

Output

For each case, output the minimum inversion number on a single line.

Sample Input

  
  
   
   10
1 3 6 9 0 8 5 7 4 2

Sample Output

Author

CHEN, Gaoli

#include<iostream>
#include<cstdio>
using namespace std ;
int a [ 5005 ],cnt ,t [ 5005 ],b [ 5005 ];
void merge_sort ( int x , int y )
{
    if (y -x > 1 )
    {
        int m =x +(y -x )/ 2 ;
        int p =x ,q =m ,i =x ;
        merge_sort (x ,m );
        merge_sort (m ,y );
        while (p <m ||q <y )
        {
               if (q >=y ||(p <m &&a [p ]<=a [q ])) t [i ++]=a [p ++];
               else {t [i ++]=a [q ++];cnt +=m -p ;}
        }
        for (i =x ;i <y ;i ++) a [i ]=t [i ];
     }

}

int main ()
{
    int n ;
    while (cin >>n )
    {
        for ( int i = 0 ;i <n ;i ++) cin >>a [i ],b [i ]=a [i ];
        cnt = 0 ;
        merge_sort ( 0 ,n );
        int minc =cnt ;
        for ( int i = 0 ;i <n ;i ++)
        {
            cnt -=b [i ];
            cnt +=n -b [i ]- 1 ;
            if (cnt <minc ) minc =cnt ;
        }
        cout <<minc <<endl ;
    }
    return 0 ;
}