Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19740 Accepted Submission(s): 11866

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

16

Author

CHEN, Gaoli

Source

题意:先输入一个n代表有n个数,然后输入n个数(0~n-1),可以把第一个数移到最后形成新的数列,问最小的逆序数是多少。

逆序数: 也就是说，对于n个不同的元素，先规定各元素之间有一个标准次序（例如n个不同的自然数，可规定从小到大为标准次序），于是在这n个元素的任一排列中，当某两个元素的先后次序与标准次序不同时，就说有1个逆序。一个排列中所有逆序总数叫做这个排列的逆序数。

(一开始读完题完全不知道逆序数是啥东西,看别人博客看了很久,结构搜了一下逆序数的概念发现可以直接暴力.)

先求出题目给出的逆序数,然后用递推关系求得将第一个数移到最后形成的新的数列的逆序数. sum=sum-a[i]+(n-a[i]-1)。

将第一个数移到最后可以看成两步：

(1):将第一个数从数列中移除,会发现逆序数减少了a[1]（当前数列的第一个数)。

(2):将第一个数加入到数列的最后,会发现此时的逆序数在(1)的基础上 +n-a[1]-1。

所以就可以得出递推关系了。

#include<stdio.h>
#include<string.h>
int a[5010],b[5010];
int main()
{
int i,j,k,n,s,sum;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
memset(b,0,sizeof(b));
s=0;
for(i=1;i<n;i++)
{
k=0;
for(j=0;j<i;j++)
{
if(a[i]<a[j])
k++;
}
b[i]=k;
s=s+k;
}
sum=s;
for(i=0;i<n;i++)
{
sum=sum-a[i]+(n-a[i]-1);
if(sum<s)
s=sum;
}
printf("%d\n",s);
}
return 0;
}

HDU1394

Minimum Inversion Number

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