HDU1394 Minimum Inversion Number(线段树)

最新推荐文章于 2019-02-20 10:22:29 发布

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本文介绍了一种算法，用于解决给定数列通过循环移位寻找逆序对数量最小值的问题。通过线段树查询初始逆序对数量，并利用特定规律循环计算不同移位情况下的逆序对变化，最终找出最小逆序对数量。

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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

/***********************************************************************************************************************
题意：给出一组数，每次可以把第一个数移到最后一位，求这个过程中逆序对数最小的对数
思路：先用线段树查询初始有多少个逆序对数，然后有以下结论（我是百度的...我不知道怎么证明= =）
如果是0到n的排列，那么如果把第一个数放到最后，对于这个数列，逆序数是减少a[i],而增加n-1-a[i]
至此就好办了 一个循环解决求出最小值
***********************************************************************************************************************/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <ctime>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define LEN 5555
int sum[LEN << 2] , a[LEN];
void PushUp(int rt) {sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}
void update(int p , int l ,int r , int rt)
{
	if(l == r)
	{
		sum[rt] ++;
		return ;
	}
	int m = (l + r) >> 1;
	if(p <= m) update(p , lson);
	else       update(p , rson);
	PushUp(rt);
}
int query(int L ,int R , int l , int r , int rt)
{
	if(L <= l && r <= R) return sum[rt];
	int m = (l + r) >> 1 , ret = 0;
	if(L <= m) ret += query(L , R , lson);
	if(R > m)  ret += query(L , R , rson);
	return ret;
}
int main()
{
	//freopen("data.in" , "r" , stdin);
	int n;
	while(~scanf("%d", &n))
	{
		int ans = 0;
		memset(sum , 0 , sizeof(sum));
		for(int i = 0 ; i < n ; i ++)
		{
			scanf("%d", &a[i]);
			ans += query(a[i] , n - 1 , 0 , n - 1 , 1);
			update(a[i] , 0 , n - 1 , 1);
		}
		int temp = ans;
		for(int i = 0 ; i < n ; i ++)
		{
			temp = temp - a[i] + n - 1 - a[i];
			ans = min(ans , temp);
		}
		printf("%d\n" , ans);
	}
	return 0;
}