Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9514    Accepted Submission(s): 5860

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

 

Author

CHEN, Gaoli

 

Source

 

Recommend

Ignatius.L

 

题目大意：

求逆序数。也就是给你一个序列，每次求逆序数，然再把第一个数放到这个序列的末尾，构成新的序列。问你这n个序列的最小的逆序数。

解题思路：

1、对于每个序列，其原来的逆序数记为 pre ， 如果当前把该序列 第一个数 a[0] 移动到尾部，那么新序列的逆序数为 pre-a[i]+(n-a[i]-1)

因为序列中比a[i]大的数有 n-a[i]-1 个，比a[i]小的有 a[i]个。

因此只需求出第一个序列的逆序数，依次可以递推出这n个序列的逆序数，求出最小的即可

2、求第一个序列的逆序数的方法

（1）暴力算法，据说不会超时

（2）线段树，建 [0,n]这段树，对于数据 a[i] ，先查询 (a[i]+1,n) 这段的值也就是比a[i]大的数的个数也就是 逆序数，然后插入 (a[i],a[i]) 值为1

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int maxn=5100;struct tree{    int l,r,sum;}a[maxn*4];int data[maxn],n,m;void build(int l,int r,int k){    a[k].l=l;    a[k].r=r;    a[k].sum=0;    if(l
         
          =mid+1) insert(l,r,2*k+1,c); else{ insert(l,mid,2*k,c); insert(mid+1,r,2*k+1,c); } a[k].sum=a[2*k].sum+a[2*k+1].sum; }}int query(int l,int r,int k){ if(l<=a[k].l && a[k].r<=r){ return a[k].sum; }else{ int mid=(a[k].l+a[k].r)/2; if(r<=mid) return query(l,r,2*k); else if(l>=mid+1) return query(l,r,2*k+1); else{ return query(l,mid,2*k) + query(mid+1,r,2*k+1) ; } }}void solve(){ int ans=0; build(1,n,1); for(int i=1;i<=n;i++){ ans+=query(data[i]+1,n,1); insert(data[i]+1,data[i]+1,1,1); //cout<
          
           <<" "<
           
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