HDU 1394 Minimum Inversion Number（树状数组+逆序数）

本文介绍了一种算法，用于寻找给定序列通过不同元素位移后得到的序列中逆序数最小的情况。采用树状数组优化求解过程，实现快速更新及区间求和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

树状数组更新单点值，求区间和。这里用到了一个数学知识：设ans为逆序和，a[i]输入的序列，则将a[i]移至队尾后，序列的逆序数变为ans=ans-a[i]+n-a[i]-1；在求出初次输入的逆序数后，进行一个 1~n for循环即可求解。代码如下：

#include<stdio.h> #include<string.h> int tree[5005],a[5005]; int n; int lowbit(int x) { return x&-x; } int query(int x) { int s=0; while(x>0) { s+=tree[x]; x-=lowbit(x); } return s; } void add(int x,int v) { while(x<=n) { tree[x]+=v; x+=lowbit(x); } } int main() { int i,x,ans; while(scanf("%d",&n)!=EOF) { memset(tree,0,sizeof(tree)); memset(a,0,sizeof(a)); ans=0; for(i=1;i<=n;++i) { scanf("%d",&a[i]); ans+=query(n)-query(a[i]+1),add(a[i]+1,1); } x=ans; for(i=1;i<=n;++i) { ans+=n-a[i]-a[i]-1; if(x>ans) x=ans; } printf("%d\n",x); } return 0; }

#include<stdio.h>
#include<string.h>
int tree[5005],a[5005];
int n;
int lowbit(int x)
{
    return x&-x;
}
int query(int x)
{
    int s=0;
    while(x>0)
    {
        s+=tree[x];
        x-=lowbit(x);
    }
    return s;
}
void add(int x,int v)
{
    while(x<=n)
    {
        tree[x]+=v;
        x+=lowbit(x);
    }
}
int main()
{
    int i,x,ans;
    while(scanf("%d",&n)!=EOF)
    {
        memset(tree,0,sizeof(tree));
        memset(a,0,sizeof(a));
        ans=0;
        for(i=1;i<=n;++i)
        {
            scanf("%d",&a[i]);
            ans+=query(n)-query(a[i]+1),add(a[i]+1,1);
        }
        x=ans;
        for(i=1;i<=n;++i)
        {
            ans+=n-a[i]-a[i]-1;
            if(x>ans) x=ans;
        }
        printf("%d\n",x);
    }
    return 0;
}