hdu1394 [Minimum Inversion Number]

Minimum Inversion Number

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.

InputThe 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. 
OutputFor 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

题目大意：　　ai是0——n-1之间的一个数字，且互不重复；　　给你一个长度为n的数列{ai}，求以其中某个数为起点的循环队列中的逆序对个数最小；　　For example：　　　　{a1,a2,a3..,an}　　　　{a2,a3..,an,a1}　　　　{a3,..an,a1,a2}　　　　{an,a1,a2,..an-1}

Tips:　　　　因为ai是0——n-1之间的一个数字，且互不重复；　　所以我们可以知道当ai为开头时，有多少数比它大，有多少数比它小；　　F[i]表示以第i个数为起点的逆序对个数；　　所以先用（线段树|树状数组|归并排序）求出原序列的逆序对数设为F[1]；　　所以不难推出F[i]=F[i-1]-a[i-1]（有a[i-1]个比它小的数）+n-a[i-1]-1（有n-a[i-1]-1个比它大的数）,即F[i]=F[i-1]+n-2*a[i-1]-1;　　最后取最小值，输出即可；

Code：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#define MAXN 200008
using namespace std;
int ans,n,m,a[MAXN],tree[MAXN],sum;

void add(int l,int r,int x,int v){
    if(l==r){
        tree[v]++;
        return;
    }
    int mid=(l+r) >> 1;
    if(x<=mid)
        add(l,mid,x,v<<1);
    else
        add(mid+1,r,x,(v<<1)+1);
    tree[v]=tree[v<<1]+tree[(v<<1)+1];
}

int query(int l,int r,int x,int y,int v){
    if(l==x&&r==y){
        return tree[v];
    }
    int mid=(l+r) >> 1;
    if(y<=mid)
        return query(l,mid,x,y,v<<1);
    if(x>mid)
        return query(mid+1,r,x,y,(v<<1)+1);
    return query(l,mid,x,mid,v<<1)+query(mid+1,r,mid+1,y,(v<<1)+1);
}

int    main(){
    while(scanf("%d",&n)!=EOF){
        memset(tree,0,sizeof(tree));
        sum=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            a[i]++;
            sum+=query(1,n,a[i],n,1);
            add(1,n,a[i],1);
        }
        ans=sum;
        for(int i=1;i<=n-1;i++){
            sum=sum-a[i]+1+n-a[i];
            ans=min(ans,sum);
        }
        printf("%d\n",ans);
    }
}