hdu 1394 Minimum Inversion Number 线段树

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

 

求逆序数的最小值

 

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=5010;
int sum[maxn<<2];
void push_up(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1 | 1];
}
void build(int rt,int first,int end)
{
     sum[rt]=0;
    if(first==end)
        return ;
    int mid=(first+end)>>1;
    build(rt<<1,first,mid);
    build(rt<<1 | 1,mid+1,end);
}
int quary(int rt,int first,int end,int left,int right)
{
    if(left<=first && right>=end)
        return sum[rt];
    int mid=(first+end)>>1;
    int ans=0;
    if(left<=mid)
        ans+=quary(rt<<1,first,mid,left,right);
    if(right>mid)
        ans+=quary(rt<<1 | 1,mid+1,end,left,right);
    return ans;
}
void update(int rt,int first,int end,int a)
{
    if(first==end)
    {
        sum[rt]++;
        return ;
    }
    int mid=(first+end)>>1;
    if(a<=mid)
        update(rt<<1,first,mid,a);
    else
        update(rt<<1 | 1,mid+1,end,a);
        push_up(rt);
}
int a[maxn];
int main()
{
    int n;
    while(cin>>n)
    {
        build(1,0,n-1);
        int sum=0;
        for(int i=0;i<n;i++)
        {
             cin>>a[i];
             sum+=quary(1,0,n-1,a[i],n-1);
             update(1,0,n-1,a[i]);
        }
        int ans=sum;
        for(int i=0;i<n;i++)
        {
            sum+=n-a[i]-a[i]-1;
            ans=min(ans,sum);
        }
        cout<<ans<<endl;
    }
    return 0;
}