hdu1394

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

-----------AC----------------

虽然数据只有5000，但暴力会超时

注意一点：当第p[i]个数据跑到最后面时，总数应该先减去这个a[i]的逆序数，sum-num[i]，然后这个a[i]排到最后面，即加上除了p[i]外比p[i]大的数目，n-num[i]-1，即总数sum=sum-num[i]+(n-num[i]-1)

模拟一下例题

             1　　2　　3　　4　　5　　6　　7　　8　　9　　10

p[i]        1　　3　　6　　9　　0　　8　　5　　7　　4　　2

num[i]   1　　2　　4　　6　　0　　4　　2　　2　　1　　0

sum=22

 

当p[1]排到后面，更新num[i]，每个比p[1]大的都加一，sum=sum-num[i]+(n-num[i]-1)

             1　　2　　3　　4　　5　　6　　7　　8　　9　　10

p[i]        3　　6　　9　　0　　8　　5　　7　　4　　2　　1

num[i]   3　　5　　7　　0　　5　　3　　3　　2　　1　　0

sum=29

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 5010
#define INF 999999999
int main()
{
    int n;
    while(cin>>n)
    {
        int p[MAXN],sum = 0,num[MAXN];
        int i,j;
        for(i = 0; i < n; ++i)
            cin>>p[i];
        memset(num,0,sizeof(num));
        for(i = 0; i < n-1; ++i)
        {
            for(j = i; j < n; ++j)
                if(p[i] > p[j])
                num[i]++;
            sum += num[i];
            cout<<num[i]<<" ";
        }
        int t = sum;
        for(i = 0; i< n-1; ++i)
        {
            t = t - num[i] + (n-num[i]-1);
            sum = min(sum,t);
 //           cout<<t<<endl;
            //更新num[i]值
            for(j = i+1; j<n; ++j)
                if(p[j] >p [i])
                num[j] ++;
        }
        cout<<sum<<endl;
    }
    return 0;
}