hdu 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5213    Accepted Submission(s): 3193

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

 

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Ignatius.L

线段树解法

#include<stdio.h>

#define MAXN 5001
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

int sum[MAXN<<2];
int x[MAXN];

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

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

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


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

int main(void)
{
    int n;
    while(~scanf("%d",&n))
    {
        build(0,n-1,1);
        int i,sum=0;
        for(i=1;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 min=sum;
        for(i=1;i<n;i++)
        {
            sum+=n-x[i]-x[i]-1;
            if(sum<min)
                min=sum;
        }
        printf("%d\n",min);
    }
    return 0;
}