hdoj-1394 Minimum Inversion Number 求最小逆序数（树状数组）

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本文探讨了给定序列通过不同位移操作后产生的序列中逆序数的最小值问题，并提供了一种有效的算法实现方案。文章详细介绍了如何利用树状数组来高效计算逆序数的变化，进而找出所有可能序列中的最小逆序数。

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Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22602 Accepted Submission(s): 13452

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

  101 3 6 9 0 8 5 7 4 2
 

Sample Output

16

题目链接

先求出逆序数
每次交换后的逆序数值可以直接算出来，
移动后当前逆序对数就要减去比第一个数小的个数，再加上比它大的数的个数
这样我们求出一次移动后的逆序对数
这时候我们发现下一次移动直接修改答案就好，不需要改动树状数组

推出式子ans+=n-reflect[i]-(reflect[i]-1)

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=500005;
struct Node
{
    int val;
    int pos;
};
Node node[N];
int c[N],reflect[N],n;
bool cmp(const Node& a, const Node& b)
{
    return a.val < b.val;
}
int lowbit(int x)
{
    return x & (-x);
}
void update(int x)
{
	  for(int i = x; i <= n; i += lowbit(i))
     {
         c[i] ++;
     }
}
int getsum(int x)
{
	 int sum = 0;
    for(int i = x; i > 0; i -= lowbit(i))
    {
        sum += c[i];
    }
    return sum;
}
int main()
{
    while (scanf("%d",&n)!=EOF&&n)
    {
        for (int i=1;i<=n;++i)
        {
            scanf("%d", &node[i].val);
            node[i].pos = i;
        }
        sort(node+1,node+n+1,cmp);
        for(int i = 1; i <= n; ++i)
            reflect[node[i].pos] = i;
        for(int i=1; i<=n;++i)
            c[i]=0;
        long long ans = 0;
        for(int i=1;i<=n;++i)
        {
            update(reflect[i]);
            ans+= i-getsum(reflect[i]);
        }
        long long mn=ans;
        mn=min(mn,ans);
        for(int i=1;i<=n;i++)
        {
            ans+=n-reflect[i]-(reflect[i]-1);
            mn=min(mn,ans);
        }
        printf("%lld\n",mn);
    }
    return 0;
}