Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4647    Accepted Submission(s): 2809


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
 
 
 
本题就是求循环移位后逆序数的最小值。
其实主要就是求序列的逆序数。
逆序数的求法很多，可以用归并排序求。
也可以用树状数组和线段树求逆序数。
 
逆序数求得之后，把第一个数移到最后的逆序数是可以直接得到的。
比如原来的逆序数是ans,把a[0]移到最后后，减少逆序数a[0]，同时增加逆序数n-a[0]-1个
就是ans-a[0]+n-a[0]-1;
 
只要i从0-n-1循环一遍取最小值就可以了。
 
 
线段树的做法：
/*
HDU 1394
G++ 78ms 280K

*/

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=5050;
struct Node
{
    int l,r;
    int sum;
}segTree[MAXN*3];
void Build(int i,int l,int r)
{
    segTree[i].l=l;
    segTree[i].r=r;
    if(l==r)
    {
        segTree[i].sum=0;
        return;
    }
    int mid=(l+r)>>1;
    Build(i<<1,l,mid);
    Build((i<<1)|1,mid+1,r);
    segTree[i].sum=0;
}
void add(int i,int t,int val)
{
    segTree[i].sum+=val;
    if(segTree[i].l==segTree[i].r)
    {
        return;
    }
    int mid=(segTree[i].l+segTree[i].r)>>1;
    if(t<=mid) add(i<<1,t,val);
    else add((i<<1)|1,t,val);
}
int sum(int i,int l,int r)
{
    if(segTree[i].l==l&&segTree[i].r==r)
      return segTree[i].sum;
    int mid=(segTree[i].l+segTree[i].r)>>1;
    if(r<=mid) return sum(i<<1,l,r);
    else if(l>mid)  return sum((i<<1)|1,l,r);
    else return sum(i<<1,l,mid)+sum((i<<1)|1,mid+1,r);
}
int a[MAXN];
int main()
{
  //  freopen("in.txt","r",stdin);
  //  freopen("out.txt","w",stdout);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        Build(1,0,n-1);
        for(int i=0;i<n;i++)
          scanf("%d",&a[i]);
        int ans=0;
        for(int i=0;i<n;i++)
        {
            ans+=sum(1,a[i],n-1);
            add(1,a[i],1);
        }
        int Min=ans;
        for(int i=0;i<n;i++)
        {
            ans-=a[i];//减少的逆序数
            ans+=n-a[i]-1;
            if(ans<Min)Min=ans;
        }
        printf("%d\n",Min);
    }
    return 0;
}

 

 

树状数组：

/*
HDU 1394
AC  G++ 46ms 252k

*/


#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=5050;
int c[MAXN];
int n;
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int val)
{
    while(i<=n)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}
int a[MAXN];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int ans=0;
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
            ans+=sum(n)-sum(a[i]);
            add(a[i],1);
        }
        int Min=ans;
        for(int i=1;i<=n;i++)
        {
            ans+=n-a[i]-(a[i]-1);
            if(ans<Min)Min=ans;
        }
        printf("%d\n",Min);
    }
    return 0;
}