HDU1394Minimum Inversion Number(树状数组)

最新推荐文章于 2020-04-07 14:23:24 发布

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最新推荐文章于 2020-04-07 14:23:24 发布

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分类专栏：树状数组

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本文探讨了在一组整数序列中，通过不同排列方式找到逆序数最小的序列的方法。详细介绍了算法实现过程和核心思路。

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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

/*
** 时间：2014-10-9
** 知识点：
** 题意：
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int c[5100],n;
void add(int a)
{
    while(a<=n)
    {
        c[a]++;
        a+=a&(-a);
    }
}
int sum(int a)
{
    int res=0;
    while(a>0)
    {
        res+=c[a];
        a-=a&(-a);
    }
    return res;
}
int main()
{
    int g[5100];
    while(cin>>n)
    {
        int i,j,a,s=0,ans=0;
        for(i=0;i<n;i++)
        {
            cin>>g[i];
            g[i]++;
        }
        memset(c,0,sizeof c);
        for(i=0;i<n;i++)
        {
            a=g[i];
            ans+=a-1-sum(a);
            add(a);
        }
        s=ans;
        for(i=0;i<n;i++){
           s+=n+1-g[i]-g[i];//除去本身g[i]这个数还有n-1个其中比g[i]大的有n-1-(g[i]-1),小的有g[i]-1
           //因为放到最后面所以g[i]的逆序数为零同时比g[i]大的数逆序数加一，即为n-1-(g[i]-1)-(g[i]-1)
           ans=min(ans,s);
        }
        cout<<ans<<endl;
    }
    return 0;
}