HLG 1250 Minimum Inversion Number

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

线段树功能：单点增减，区间求和。

#include<stdio.h>
int v[5001];
typedef struct 
{
    int left;
    int right;
    int mid;
    int tot;
}line;
line l[20004];
void creat(int a,int b,int r)
{
    l[r].left=a;
    l[r].right=b;
    l[r].tot=0;
    l[r].mid=(a+b)>>1;
    if(a==b)
        return;
    creat(a,l[r].mid,r<<1);
    creat(l[r].mid+1,b,(r<<1)+1);
}
void update(int a,int b,int r)
{
    if(l[r].left>b||l[r].right<a)
        return;
    if(l[r].left>=a&&l[r].right<=b)
    {
        l[r].tot+=1;
        return;
    }
    update(a,b,r<<1);
    update(a,b,(r<<1)+1);
    l[r].tot=l[r<<1].tot+l[(r<<1)+1].tot;
}
int query(int a,int b,int r)
{
    if(l[r].left>b||l[r].right<a)
        return 0;
    if(l[r].left>=a&&l[r].right<=b)
        return l[r].tot;
    return  query(a,b,r<<1)+query(a,b,(r<<1)+1);
}
int main()
{
    int res,sum,n,i;
    while(scanf("%d",&n)!=EOF)
    {
        creat(1,n,1);
        sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&v[i]);
                v[i]++;
            update(v[i],v[i],1);
            sum+=query(v[i]+1,n,1);
        }
        res=sum;
        for(i=0;i<n;i++)
        {
            sum+=n-v[i]-v[i]+1;
            res=res<sum?res:sum;
        }
        printf("%d\n",res);
    }
    return 0;
}
 

方法二：树状数组

#include<stdio.h>
#include<string.h>
#define min(a,b)(a)<(b)?(a):(b);
int v[5002];
int l[5002];
int n;
int lowbit(int x)
{
    return (x)&(-x);
}
void update(int pos)
{
    while(pos<=n)
    {
        l[pos]+=1;
        pos+=lowbit(pos);
    }
}
int getsum(int x)
{
    int sum=0;
    while(x>0)
    {
        sum+=l[x];
        x-=lowbit(x);
    }
    return sum;
}
int main()
{
    int res,p,i,sum;
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        memset(l,0,sizeof(l));
        for(i=0;i<n;i++)
        {
            scanf("%d",&v[i]);
            v[i]++;
            sum+=(getsum(n)-getsum(v[i]));
            update(v[i]);
        }
        res=sum;
        for(i=0;i<n;i++)
        {
            sum+=n-v[i]-v[i]+1;
            res=min(res,sum);
        }
        printf("%d\n",res);
    }
    return 0;
}