hdu 1394 Minimum Inversion Number 最小逆序数

Minimum Inversion Number

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

Total Submission(s): 15355    Accepted Submission(s): 9366


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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#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

const int  maxn=5000;
int n;
int a[maxn+20];
int sum[4*maxn+20];
int tot;

void update(int num,int le,int ri,int x)
{
if(le==ri)
{
sum[num]++;
return;
}
int mid=(le+ri)>>1;
if(x<=mid)   update(num*2,le,mid,x);
else   update(num*2+1,mid+1,ri,x);
sum[num]=sum[num<<1]+sum[(num<<1)+1];

}

int query(int num,int le,int ri,int x,int y)
{
if(x<=le&&ri<=y)
{
return sum[num];
}
int ret=0;
int mid=(le+ri)>>1;
if(x<=mid)   ret+=query(num*2  ,le   ,mid,x,y);
if(y> mid)   ret+=query(num*2+1,mid+1,ri ,x,y);
return ret;

}

int main()
{
while(~scanf("%d",&n))
{
memset(sum,0,sizeof sum);
tot=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
tot+=query (1,0,maxn,a[i]+1,maxn);
update(1,0,maxn,a[i]);
}
int mini=tot;
for(int i=1;i<=n-1;i++)
{
int sm=query(1,0,maxn,0,a[i])-1;
int la=n-1-sm;
tot=tot-sm+la;
mini=min(mini,tot);
}
printf("%d\n",mini);
}
return 0;

}