Minimum Inversion Number 数状数组

Minimum Inversion Number

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1394

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

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int n,c[5050];

int lowbit(int x)
{
    return x&(-x);
}

void add(int i,int val)
{
    while(i<=n)
    {
        c[i]=c[i]+val;
        i=i+lowbit(i);
    }
}

int sum(int i)
{
    int s=0;
    while(i)
    {
        s=s+c[i];
        i=i-lowbit(i);
    }
    return s;
}

int main()
{
    int i,j,k;
    int a[5050];
    while(scanf("%d",&n)!=EOF)
    {
        int cnt=0;
        memset(c,0,sizeof(c));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
            cnt=cnt+sum(n)-sum(a[i]);
            add(a[i],1);
        }
        int mi=cnt;
        for(i=1;i<n;i++)
        {
            cnt=cnt-(a[i]-1)+(n-a[i]);
            if(cnt<mi)
                mi=cnt;
        }
        printf("%d
",mi);
    }
    return 0;
}