hdu 1394 Minimum Inversion Number（这道题改日我要用线段树再做一次哟~）

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

 

 

//有的大神是用线段树解的，原谅我怎么都无法看出线段树来。。。。_(:з」∠)_

//于是采用暴力解，然而。。超时了   (╯‵□′)╯︵┴─┴

//于是借鉴别人的代码~~~  (●'◡'●)

//思路：每次把末尾的数掉到序列前面时，减少的逆序对数为n-1-a[i] ，

//增加的逆序对数为a[i] ，这样就可在所有的序列中找出含有逆序对最少的了！（ps：给跪了，这是性质吗。。。

 

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

int main()
{
    int n,ans,k;
    int data[];
    while(cin>>n)
    {
        ans=;
        for(int i=;i<n;i++)
        scanf("%d",&data[i]);
        for(int i=;i<n;i++)
        {
            for(int j=i;j<n;j++)
            {
                if(data[i]>data[j])
                ans++;
            }
        }
        k=ans;
        for(int i=n-;i>=;i--)
        {
            k-=n--data[i];
            k+=data[i];
            if(ans>k)
            ans=k;
        }
        cout<<ans<<endl;
    }
    return ;
}

//再贴上我的狗血超时代码。。。

 

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

int find(int a[],int n)
{
    int ans=;
    for(int i=;i<n;i++)
    {
        for(int j=i+;j<n;j++)
        {
            if(a[i]>a[j])
            ans++;
        }
    }
    return ans;
}

int main()
{
    int n;
    int data[],num[];
    while(scanf("%d",&n)!=-)
    {
        for(int i=;i<n;i++)
        scanf("%d",&data[i]);
        for(int i=;i<n;i++)
        {
            num[i]=find(data,n);
            int tmp=data[n-];
            for(int j=n-;j>=;j--)
            {
                data[j]=data[j-];
            }
            data[]=tmp;
        }
        sort(num,num+n);
        cout<<num[]<<endl;
    }
    return ;
}