Minimum Inversion Number


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18379    Accepted Submission(s): 11158


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​​


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题解：

问你不断地将开头的数放到结尾。求其中最小的逆序数。



树状数组求出最初的逆序数。复杂度O（nlogn），然后O(n)查询其他解。



AC代码：



#pragma comment(linker, "/STACK:102400000,102400000")
//#include<bits/stdc++.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <bitset>
#include <iomanip>
#include <list>
#include <stack>
#include <utility> 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
const double eps = 1e-8;  
const int INF = 1e9+7; 
const ll inf =(1LL<<62) ;
const int MOD = 1e9 + 7;  
const ll mod = (1LL<<32);
const int N =1e6+6; 
const int M=100010;
const int maxn=5010;
#define mst(a) memset(a, 0, sizeof(a))
#define M_P(x,y) make_pair(x,y)
#define in freopen("in.txt","r",stdin) 
#define rep(i,j,k) for (int i = j; i <= k; i++)  
#define per(i,j,k) for (int i = j; i >= k; i--)  
#define lson  l , mid , rt << 1    
#define rson  mid + 1 , r , rt << 1 | 1  
const int lowbit(int x) { return x&-x; }
//const int lowbit(int x) { return ((x)&((x)^((x)-1))); } 
int read(){ int v = 0, f = 1;char c =getchar();
while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}
while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();
return v*f;}
int a[maxn];
int n,m;
int c[maxn];
void add(int i,int val)
{
    while(i<=n)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int s = 0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}
int main()
{
    
    while(~scanf("%d",&n))
    {
        int ans = 0;
        memset(c,0,sizeof(c)); 
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
            ans+=sum(n)-sum(a[i]);
            add(a[i],1);
        }
        int MIN = ans;
        for(int i=1;i<=n;i++)
        {
            ans-=a[i];//减少的逆序数 
            ans+=n-a[i]+1; //增加的逆序数 
            if(ans<MIN) MIN=ans;
        }
        cout<<MIN<<endl;
    }
    return 0;
}