Minimum Inversion Number - HDU 1394 树状数组

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本文探讨了如何通过树状数组或线段树算法解决最小逆序数问题，即找到一系列数字排列中逆序对数量最少的情况。文章提供了完整的代码实现，并解释了核心算法思想。

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11249 Accepted Submission(s): 6912

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

题意：每次求前面比后面大的对数，然后改变原序列顺序，取其中的最小值为答案。

思路：先用树状数组求出当前序列的对数，然后ret=ret-(num[i]-1)+(n-num[i]);线段树也可以。

树状数组AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,tree[5010],num[5010];
int lowbit(int x)
{
    return x&(-x);
}
void update(int pos)
{
    for(;pos<=n;pos+=lowbit(pos))
       tree[pos]++;
}
int query(int pos)
{
    int ret=0;
    for(;pos>0;pos-=lowbit(pos))
       ret+=tree[pos];
    return ret;
}

int main()
{
    int i,j,k,ans,ret;
    while(~scanf("%d",&n))
    {
        memset(tree,0,sizeof(tree));
        memset(num,0,sizeof(num));
        ret=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&k);
            k++;
            update(k);
            num[i]=k;
            ret+=i-1-query(k-1);
        }
        ans=ret;
        for(i=1;i<=n;i++)
        {
            ret=ret-(num[i]-1)+(n-num[i]);
            ans=min(ans,ret);
        }
        printf("%d\n",ans);
    }
}

线段树AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int num,l,r;
}tree[20010];
int val[5010];
void build(int o,int l,int r)
{
    tree[o].l=l;
    tree[o].r=r;
    tree[o].num=0;
    if(l==r)
      return;
    int mi=(l+r)>>1;
    build(o<<1,l,mi);
    build(o<<1|1,mi+1,r);
}
void update(int o,int pos)
{
    if(tree[o].l==pos && tree[o].r==pos)
    {
        tree[o].num=1;
        return;
    }
    int mi=(tree[o].l+tree[o].r)>>1;
    if(pos<=mi)
      update(o<<1,pos);
    else
      update(o<<1|1,pos);
    tree[o].num=tree[o<<1].num+tree[o<<1|1].num;
}
int query(int o,int l,int r)
{
    if(tree[o].l==l && tree[o].r==r)
      return tree[o].num;
    int mi=(tree[o].l+tree[o].r)>>1;
    if(r<=mi)
      return query(o<<1,l,r);
    if(l>mi)
      return query(o<<1|1,l,r);
    return query(o<<1,l,mi)+query(o<<1|1,mi+1,r);
}
int main()
{
    int n,i,j,k,ret,ans;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&val[i]);
            val[i]++;
        }
        build(1,1,n);
        ret=0;
        for(i=1;i<=n;i++)
        {
            ret+=query(1,val[i],n);
            //printf("%d\n",ret);
            update(1,val[i]);
        }
        ans=ret;
        //printf("%d\n",ret);
        for(i=1;i<=n;i++)
        {
            ret=ret-(val[i]-1)+(n-val[i]);
            //printf("%d\n",ret);
            ans=min(ans,ret);
        }
        printf("%d\n",ans);
    }
}