hdu1394Minimum 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

看了别人的代码做出来了。。 这题先用线段树的方法把读入数据的逆序值求出来，依次插入点，每插入一个点k就看[k+1,n]中前面插入的点有几个，即为该点的逆序值，然后再更新从根线段到[k,k]这些线段的值。点都插入完毕后，再一次把最前面的点移到最后面，此时总逆序数会减少a[i]-1,增加n-a[i]，然后求出最小的逆序数就行了。

#include<stdio.h>
#include<string.h>
#define maxn 5005
int a[maxn];
struct node
{
    int l,r,sum;
}b[4*maxn];
void build(int l,int r,int i)
{
    int mid;
    b[i].l=l;
    b[i].r=r;
    b[i].sum=0;
    if(l==r)
    return;
    mid=(l+r)/2;
    build(l,mid,i*2);
    build(mid+1,r,i*2+1);
}

int inverse(int l,int r,int i)
{
    int mid;
    if(b[i].l==l && b[i].r==r)
    {
        return b[i].sum;
    }
    mid=(b[i].l+b[i].r)/2;
    if(l>mid)
    return inverse(l,r,i*2+1);
    else if(r<=mid)
    return inverse(l,r,i*2);
    else if(r>mid && l<=mid)
    return inverse(l,mid,i*2)+inverse(mid+1,r,i*2+1);
}

void change(int i,int id)
{
    int mid;
    if(b[i].l==id && b[i].r==id)
    {
        b[i].sum=1;
        return;
    }
    mid=(b[i].l+b[i].r)/2;
    if(id<=mid)
    change(i*2,id);
    else if(id>mid)
    change(i*2+1,id);
    b[i].sum=b[i*2].sum+b[i*2+1].sum;
}

int main()
{
    int n,m,i,j,ans,min;
    while(scanf("%d",&n)!=EOF)
    {
        build(1,n,1);
        ans=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            ans=ans+inverse(a[i]+1,n,1);
            change(1,a[i]);
        }
        min=ans;
        for(i=1;i<=n;i++)
        {
            ans=ans+n-2*a[i]-1;
            if(ans<min)
            min=ans;
        }
        printf("%d
",min);
    }
}