HDUOJ---1754 Minimum Inversion Number （单点更新之求逆序数）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9342    Accepted Submission(s): 5739

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

ZOJ Monthly, January 2003

 

代码：

 //线段树实现单点更新，并求和
 #include<stdio.h>
 #define maxn 5001
 struct node{
   int lef,rig,sum;
   int mid(){ return lef+((rig-lef)>>) ;}
 };
  node seg[maxn<<];
  int aa[maxn+];
 void build(int left,int right,int p )
 {
     seg[p].lef=left;
     seg[p].rig=right;
     seg[p].sum=;
     if(left==right) return ;
     int mid=seg[p].mid();
     build(left,mid,p<<);
     build(mid+,right,p<<|);
 }
 void updata(int pos,int p,int val)
 {
     if(seg[p].lef==seg[p].rig)
     {
       seg[p].sum+=val;
       return ;
     }
     int mid=seg[p].mid();
     if(pos<=mid) updata(pos,p<<,val);
     else updata(pos,p<<|,val);
    seg[p].sum=seg[p<<].sum+seg[p<<|].sum;
 }
 int query(int be ,int en,int p)
 {
     if(be<=seg[p].lef&&seg[p].rig<=en)
         return seg[p].sum;
     int mid=seg[p].mid();
     int res=;
     if(be<=mid) res+=query(be ,en ,p<<);
     if(mid<en)  res+=query(be ,en ,p<<|);
     return res;
 }
 int main()
 {
     int nn,i,ans;
     while(scanf("%d",&nn)!=EOF)
     {
       ans=;
       build(,nn-,);
      for(i=;i<=nn;i++)
      {
        scanf("%d",&aa[i]);
        updata(aa[i],,);
        if(aa[i]!=nn-) ans+=query(aa[i]+,nn-,); //统计比其大的数
      }
      int min=ans;
      for(i=;i<=nn;i++)
      {
        ans+=nn-*aa[i]-;
        if(min>ans) min=ans;
      }
      printf("%d\n",min);
     }
     return  ;
 }