HDU1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5950    Accepted Submission(s): 3625

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

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

Ignatius.L

 

 

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=5010;

#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)

struct Tree{
    int l,r;
    int sum;
}tree[N<<2];

void PushUp(int rt){
    tree[rt].sum=tree[L(rt)].sum+tree[R(rt)].sum;
}

void build(int L,int R,int rt){
    tree[rt].l=L;
    tree[rt].r=R;
    if(tree[rt].l==tree[rt].r){
        tree[rt].sum=0;
        return ;
    }
    int mid=(L+R)>>1;
    build(L,mid,L(rt));
    build(mid+1,R,R(rt));
    PushUp(rt);
}

void update(int id,int rt){
    if(tree[rt].l==tree[rt].r){
        tree[rt].sum++;
        return ;
    }
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(id<=mid)
        update(id,L(rt));
    else if(id>=mid+1)
        update(id,R(rt));
    PushUp(rt);
}

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

int num[N];

int main(){

    //freopen("input.txt","r",stdin);

    int n;
    while(~scanf("%d",&n)){
        build(0,n-1,1);
        int tmp=0;
        for(int i=0;i<n;i++){
            scanf("%d",&num[i]);
            tmp+=query(num[i],n-1,1);
            update(num[i],1);
        }
        int res=tmp;
        for(int i=0;i<n;i++){
            tmp+=(n-1-num[i])-num[i];
            res=min(res,tmp);
        }
        printf("%d\n",res);
    }
    return 0;
}