Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12141Accepted Submission(s): 7413

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,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,an-1,an (where m = 0 - the initial seqence)
a2,a3,a1 (where m = 1)
a3,a4,a1,a2 (where m = 2)
...
an,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

题意是:

给N个数(0~N-1)组成的序列,求出它的逆序数(就是由多少对数,前一个比后一个大),

然后每次把第一个数放到最后,再求逆序数,然后输出所有逆序数中最小的那个...

思路:

第1步,向数组中依次存入初始序列,同时用线段树求出这个数贡献的的逆序数,就是查询比这个数大的数有几个,及用查询线段树 [x[i],N-1] 这个范围

具体方法是:叶子节点只存储两个值0和1,0表示该叶子节点表示的数还没有存入,1表示该叶子节点表示的数已经存入了,非叶子节点为两个子节点的和,表示该范围内有几个数已经存入了数组;数组中每存入一个数,先去查询比它大的数有几个,就是它贡献的逆序数,加入sum,然后更新该点,表示已经放入了;最终sum就是初始序列的逆序数

第2步,每次把第一个数放到最后,因为知道了前一个序列的逆序数,就可以用O(1)的方法求出它的逆序数

具体方法是:

比如初始序列是1 3 6 9 0 8 5 7 4 2,根据第1步,求出了它的逆序数sum

现在要把1移到最后,构成3 6 9 0 8 5 7 4 2 1,求该序列的逆序数

在初始序列中,1贡献的逆序数是排在它后面的,所有比它小的数的个数(只有0,1个),由取走1,sum先减去1;放到最后,能增加的逆序数,要往前看,是所有比1大的数的个数,是N-1-1个,所以sum=(sum-1)+(N-1-1)

然后逸动3,sum先减去3(比3小的是0,1,2),移到最后,sum加上N-1-3

因为初始序列是存储在数组中的,所以用一个for循环,每次移动x[i],比它小的有x[i]个,比它大的有N-1-x[i]个,所以sum=(sum-x[i])+(N-1-x[i])

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

//#define lson left,mid,index<<1
//#define rson mid+1,right,index<<1|1

const int MAXN=5010;
int num[MAXN<<2];

void PushUp(int ret)
{
    num[ret]=num[ret<<1]+num[ret<<1|1];
}

void BuildTree(int left,int right,int ret)
{
    num[ret]=0;
    if(left==right)
        return ;
    int mid=(left+right)>>1;
    BuildTree(left,ret<<1);
    BuildTree(mid+1,ret<<1|1);
}

void Update(int target,int left,int ret)
{
    if(left==right)
    {
        num[ret]++;
        return ;
    }
    int mid=(left+right)>>1;
    if(target<=mid)
        Update(target,left,ret<<1);
    else
        Update(target,mid+1,ret<<1|1);
    PushUp(ret);
}

/*
int Query(int L,int R,int ret)
{
	if(L<=left && right<=R)
		return num[index];
	int mid=(left+right)>>1;
	int ans=0;
	if(L<=mid)
		ans+=Query(L,R,lson);
	if(R>mid)
		ans+=Query(L,rson);
	return ans;
}
*/
int Query(int L,int ret)
{
    if(L<=left && right<=R)
        return num[ret];
    int mid=(left+right)>>1;
    if(R<=mid)
        return Query(L,ret<<1);
    else if(mid<L)
        return Query(L,ret<<1|1);
    else
        return Query(L,ret<<1)+Query(L,ret<<1|1);
}

int main()
{
    freopen("1394.in","r",stdin);
//     freopen("1394.out","w",stdout);
    int n;
    int a[5100];
    while(scanf("%d",&n)!=EOF)
    {
        int sum=0;
        BuildTree(0,n-1,1);
        for(int i=0;i<n;i++)
        {
            scanf("%d",a+i);
            sum+=Query(a[i],1);
            Update(a[i],1);
        }
        int ans=sum;
        for(int i=0;i<n;i++)
        {
            sum+=n-a[i]-1-a[i];
            ans=min(ans,sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}

HDU—1394—Minimum_Inversion_Number—【数据结构】【线段树】【单点更新】

Minimum Inversion Number

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