HDU 1394 Minimum Inversion Number 线段树

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本文介绍了一种使用线段树来求解一系列排列中最小逆序数的方法。通过不断将序列首元素移至末尾形成新序列，利用线段树高效更新并计算每个新序列的逆序数，最终找到所有可能序列中的最小逆序数。

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Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17537 Accepted Submission(s): 10640

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

用线段树求数列的逆序数
题目大意，从初始的数组开始，每次把第一个数加到最后一个，求逆序数，共求出n个逆序数，找出最小的逆序数
当求出初始数列的逆序数之后其他数列的逆序数的规律是：sum = sum + (n - 1 - a[i]) - a[i];

//hdu 1394 Minimum Inversion Number 线段树 
#include <iostream>
#include <stdio.h>
#include <string.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn=200002;
int sum[maxn<<2],a[maxn];

void pushup(int rt)
{
	sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
	sum[rt]=0;
	if(l==r) return ;
	int m=(l+r)>>1;
	build(lson);
	build(rson);
	pushup(rt);
}

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);
	if(p>m) 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 ans;
	int m=(l+r)>>1;
	ans=0;
	if(L<=m) ans+=query(L,R,lson);
	if(R>m)  ans+=query(L,R,rson);
	return ans;
}

int main()
{
	std::cin.tie(false);
    ios::sync_with_stdio(false);
	int i,j,n;
	while(~scanf("%d",&n))
	{
		int Sum=0;
		build(1,n,1);
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			Sum+=query(a[i],n-1,0,n-1,1);
			update(a[i],0,n-1,1);
		}
		int ret=Sum;
		for(int i=0;i<n;i++)
		{
			Sum=Sum+n-2*a[i]-1;
			if(ret>Sum) ret=Sum;
		}
		printf("%d\n",ret);
	}
}