hdu 1394 Minimum Inversion Number 归并排序求逆序数 求一串数的最小逆序数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 927    Accepted Submission(s): 514

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

 

#include<iostream>
#include<cstdio>
using namespace std;
int a[5005],cnt,t[5005],b[5005];
void merge_sort(int x,int y)
{   
    if(y-x>1)   
    {       
        int m=x+(y-x)/2;       
        int p=x,q=m,i=x;   
        merge_sort(x,m);   
        merge_sort(m,y);   
        while(p<m||q<y)      
        {        
               if(q>=y||(p<m&&a[p]<=a[q])) t[i++]=a[p++];        
               else {t[i++]=a[q++];cnt+=m-p;}    
        }     
        for(i=x;i<y;i++) a[i]=t[i];  
     }
    
}

int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=0;i<n;i++) cin>>a[i],b[i]=a[i];
        cnt=0;
        merge_sort(0,n);
        int minc=cnt;
        for(int i=0;i<n;i++)
        {
            cnt-=b[i];
            cnt+=n-b[i]-1;
            if(cnt<minc) minc=cnt;
        }
        cout<<minc<<endl;
    }
    return 0;
}