Minimum Inversion Number

最新推荐文章于 2018-11-03 20:59:41 发布

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本文介绍了一个关于求解序列通过不同轮转方式得到的多个序列中具有最小逆序数的问题，并提供了一段C++代码实现，用于解决此类问题。通过对输入序列的不同轮转，计算每种情况下序列的逆序数，并找到逆序数最小的情况。

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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

Author

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define MAX 5050
using namespace std;
struct Node
{
    int l,r,sum;
}node[MAX<<2];
void build_tree(int l,int r,int root)
{
    node[root].l=l;
    node[root].r=r;
    if(l==r)
    {
        node[root].sum=0;
        return ;
    }
    int mid=(l+r)>>1;
    build_tree(l,mid,root<<1);
    build_tree(mid+1,r,(root<<1)|1);
    node[root].sum=0;
}
void add(int i,int t,int val)
{
    node[i].sum+=val;
    if(node[i].l==node[i].r){
        return ;
    }
    int mid=(node[i].l+node[i].r)>>1;
    if(t<=mid)add(i<<1,t,val);
    else add((i<<1)|1,t,val);
}
int sum(int l,int r,int root)
{
    if(node[root].l==l&&node[root].r==r)
        return node[root].sum;
    int mid=(node[root].r+node[root].l)>>1;
    if(r<=mid)return sum(l,r,root<<1);
    else if(l>mid)return sum(l,r,(root<<1)|1);
    else
    {
        return sum(l,mid,root<<1)+sum(mid+1,r,(root<<1)|1);
    }
}
int a[MAX];
int main()
{
  int n;
  while(~scanf("%d",&n))
  {
      build_tree(1,n,1);
      for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
      int ans=0;
      for(int i=1;i<=n;i++)
      {
          ans+=sum(a[i]+1,n,1);
          add(1,a[i]+1,1);
      }
      int Min=ans;
      for(int i=1;i<=n;i++){
        ans-=a[i];
        ans+=n-a[i]-1;
        if(ans<Min)Min=ans;
      }
      printf("%d\n",Min);
  }
  return 0;
}