HDU_1394_Minimum Inversion Number_线段树求逆序数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16686    Accepted Submission(s): 10145

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

 

线段树求逆序数，手撸。

先求初始序列的逆序数（nlogn），再推出所有情况，求的最小值。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define maxn 5005
struct Node
{
    int l,r;
    int sum;
} tree[maxn<<2];

void build(int l,int r,int rt)
{
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].sum=0;
    if(tree[rt].l==tree[rt].r)
        return;
    int mid=(l+r)/2;
    build(lson);
    build(rson);
}

void update(int x,int l,int r,int rt)
{
    if(tree[rt].l==x&&tree[rt].r==x)
    {
        tree[rt].sum++;
        return;
    }
    int mid=(l+r)/2;
    if(x<=mid)
        update(x,lson);
    else
        update(x,rson);
    tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
}

int ans=0;
int query(int L,int R,int l,int r,int rt)
{
    if(L>R)
        return 0;
    if(L==tree[rt].l&&R==tree[rt].r)
    {
       // cout<<tree[rt].sum<<endl;
       return tree[rt].sum;
    }
    int mid=(l+r)/2;
    if(R<=mid)
        return query(L,R,lson);
    else if(L>mid)
        return query(L,R,rson);
    else
    {
        return query(L,mid,lson)+query(mid+1,R,rson);
    }
}

int num[5005];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;build(1,n,1);
        for(int i=0; i<n; i++)
        {
            scanf("%d",&num[i]);
            num[i]++;
        }
        //cout<<tree[16].l<<'*'<<tree[16].r<<endl;
        for(int i=n-1;i>=0;i--)
        {
            //cout<<num[i]<<'&'<<endl;
            if(i!=n-1)
                ans+=query(1,num[i]-1,1,n,1);
            //cout<<ans<<endl;
            update(num[i],1,n,1);
            //cout<<ans<<endl;
        }
        int res=ans;
        for(int i=0;i<n-1;i++)
        {
            res=res+n-num[i]-num[i]+1;
            ans=min(res,ans);
        }
        printf("%d
",ans);
    }
    return 0;
}