hdu1394——Minimum Inversion Number

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


树状数组求逆序数，也可以线段树和归并排序

1.求逆序数，
2循环移动后，各个序列之间逆序的关系

比如没移动的时候，有x个，移动a[0]到最后，那么现在就是     x-a[0]+n-a[0]-1  （因为这些数的大小是0~n-1，所以第一个数产生的逆序数就是这个数本身的大小，那么移动后新产生的逆序数就是n-1-这个数了

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int tree[5010];
int val[5010];
int n;
inline int lowbit(int x)
{
	return x&(-x);
}

void add(int x,int val)
{
	for(int i=x;i<=5010;i+=lowbit(i))
		tree[i]+=val;
}

int sum(int x)
{
	int ans=0;
	for(int i=x;i;i-=lowbit(i))
		ans+=tree[i];
	return ans;
}


int main()
{
	while(~scanf("%d",&n))
	{
		int ans=0,cnt=0;
		memset(tree,0,sizeof(tree));
		for(int i=1;i<=n;i++)
			scanf("%d",&val[i]);
		for(int i=1;i<=n;i++)
		{
			add(val[i]+1,1);
			ans+=i-sum(val[i]+1);
		}
		cnt=ans;
		for(int i=1;i<=n-1;i++)
		{
			cnt=cnt-val[i]+n-val[i]-1;
			ans=min(cnt,ans);
		}
		printf("%d\n",ans);
	}
	return 0;
}