HDU1394 Minimum Inversion Number

题目链接：HDU1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16503    Accepted Submission(s): 10039

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
   
   
   
   

 

题意：给出从0到n-1这n个数的一个排列，从这个顺序开始，每次都把开头的一个移到最后一个，问这些组合中最逆序数最少是多少。

题目分析：首先我们可以利用线段树求出当前这一组数的逆序数。开始构建线段树的时候令sum数组全为0，每录入一个数x[i]先统计从x[i]到n－1中sum数组的和，然后在sum中对应这个数的位置＋1，这样我们每次可以统计出来x[i]是之前多少个数的逆序数，加起来就是整个序列的逆序数。之后需要用一步O（1）的方法统计出其他解，从0到n-1依次移动x[i]到最后一位，由于x[i]已经是第一位，而且这n个数确定是从0到n－1，因此可以肯定有x[i]个数（0到x[i]-1）是原本的逆序数变成顺序的了，那其余的n-1-x[i]肯定是反过来了。因此记上一个序列的逆序数是sum的话，下一个逆序数是sum+n-x[i]-x[i]-1。然后取最小值就好了。

//
//  main.cpp
//  HDU1394
//
//  Created by teddywang on 16/4/29.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int sum[200010];
int n;
int x[5555];

void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void update(int x,int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]++;
        return ;
    }
    int m=(l+r)>>1;
    if(m>=x)
        update(x,lson);
    else update(x,rson);
    pushup(rt);
}

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

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

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        build(0,n-1,1);
        int sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x[i]);
            sum+=query(x[i],n-1,0,n-1,1);
            update(x[i],0,n-1,1);
        }
        int ans=sum;
        for(int i=0;i<n;i++)
        {
            sum+=n-x[i]-x[i]-1;
            ans=min(sum,ans);
        }
        printf("%d\n",ans);
    }
}