hdoj_1394Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6642    Accepted Submission(s): 4051

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

#include<iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#pragma warning(disable : 4996)
const int MAXN = 5005;
int tree[MAXN];
int num[MAXN];
int n;

int lowbit(int x)
{
	return x & (-x); 
}

int getsum(int pos)
{
	int sum = 0;
	while (pos > 0)
	{
		sum += tree[pos];
		pos -= lowbit(pos);
	}
	return sum;
}

void update(int pos, int value)
{
	while (pos <= n)
	{
		tree[pos] += value;
		pos += lowbit(pos);
	}
}
int main()
{
	freopen("in.txt", "r", stdin);
	int sum, ans;
	while (scanf("%d", &n) != EOF)
	{
		sum = 0;
		memset(tree, 0, sizeof(tree));
		for(int i = 1; i <= n; i++)
		{
			scanf("%d", &num[i]);
			update(num[i] + 1, 1);
			sum += getsum(n) - getsum(num[i] + 1);
			//sum += i - getsum(num[i] + 1);
		}
		//cout << sum << endl;
		ans = sum;
		for (int i = 1; i <= n; i++)
		{
			sum = sum - 2 * num[i] + n - 1;
			if(sum < ans)
			{
				ans = sum;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}