HDU1394：Minimum Inversion Number

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

题目大意：给你一个序列，每次将这个序列的最后一个元素移到第一位，问这些过程中，所生成的每种序列，最小的逆序对数为多少，输出这个最小逆序对数。

解题思路：先用树状数组求出原始序列的逆序对数，然后开始把最后一位a[n]移到第一位，逆序对增加a[i]-1，逆序对减少n-a[i]（这道题给出序列元素值的范围是0~n-1），然后每次移动更新ans就行。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int n; 
int a[500001];//初始数组 
int c[500001];
int lowbit(int x)
{
	return x&-x;
}
void add(int x)
{
	while(x<=n)
	{
		c[x]+=1;
    	x+=lowbit(x);
	}
}
int sum(int x)
{
	int s=0;
	while(x>0)
	{
		s+=c[x];
		x-=lowbit(x);
	}
	return s;
 } 
int main()
{
	while(scanf("%d",&n)!=EOF)
	{ 
	    int ans=0;
		memset(c,0,sizeof(c));
	  for(int i=1;i<=n;i++)
	  {
	  	scanf("%d",&a[i] );
	     	a[i]+=1; //树状数组不能出现0  否则在上面void add()中x+=lowbit(x)一直是0  会陷入死循环 
		  	add(a[i]);
		 	ans+=(i-sum(a[i]));//i为当前插入的个数  sum(a[i])为当前比它小的个数 
	  }
	   int  minn=50000000;
	  for(int i=n;i>0;i--)
	  {
	  	minn=min(minn,ans+a[i]-1-(n-a[i] ));//把一个数从末尾移除 则前面比它大的元素（一共有n-a[i]个）的逆序列数都减一   把该数放到首位后它本来是没有逆序列 现在比他小的（一共a[i]-1个）都能和他构成一个逆序列  
	  	ans=ans+a[i]-1-(n-a[i]);
	  }
	  printf("%d\n",minn);
	}
	return 0;
}