hdu-1394Minimum Inversion Number

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

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

 

题目大意：

输入一个整数n,一个数列x[n]，数列是由0~n-1这n个数组成的。该数列可以有多中变换。变换的规则是：每次变换只能移一次，只能将第一个数字移动到数列最后。求出每种变换之后的逆序数（包括原数列），输出这些逆序数中的最小值。

解题思路：

根据逆序数的性质，可以先求出原数列的逆序数，再递推出变换之后的逆序数。

例如：原数列的逆序数为sum,根据逆序数的性质，变换一次之后的逆序数就为：sum+(n-2*x[i]-1)

推导过程： 假设现在的数列为 a1,a2, a3,......an。此时的逆序数为sum。 

如果在a1后面的数中有，比a1大的有max个，比a1小的有min个，易知，max=n-a1-1，min=n-1-max。

当进行一次变换之后，数列变成了a2,a3,......an,a1。那么之前在a1后面的数，全部到了a1的前面，所以此时的逆序数为sum+max-min。即sum+n-2*a1-1。推广到一般形式就是：sum+n-2*x[i]-1  (sum是变换前一次的逆序数)。

解题方法：

先用线段树求出原数列的逆序数，在用公式推导出每种形式的逆序数，最后求出最小值。

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>using namespace std;#define maxn 5555#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1int sum[maxn<<2];void push(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){    sum[rt]=0;    if(l==r)        return ;    int m=(r+l)>>1;    build(lson);    build(rson);}void updata(int p,int l,int r,int rt){    if(l==r)    {        sum[rt]++;        return ;    }    int m=(r+l)>>1;    if(p<=m)        updata(p,lson);    else        updata(p,rson);    push(rt);}int query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)    {        return sum[rt];    }    int m=(r+l)>>1;    int ans=0;    if(L<=m)        ans+=query(L,R,lson);    if(R>m)        ans+=query(L,R,rson);    return ans;}int x[maxn<<1];int main(){    int n,sum;    while(~scanf("%d",&n))    {        build(0,n-1,1);        sum=0;        for(int i=0;i<n;i++)//求出原数列的逆序数sum        {            scanf("%d",&x[i]);            sum+=query(x[i],n-1,0,n-1,1);            updata(x[i],0,n-1,1);        }        int ans=sum;        for(int i=0;i<n-1;i++)//推出每种形式的逆序数，再进行比较，将较小的逆序数存在ans中        {            sum+=n-2*x[i]-1;            ans=min(ans,sum);        }        printf("%d\n",ans);    }    return 0;}