学步园

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.

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.

For each case, output the minimum inversion number on a single line.

10
1 3 6 9 0 8 5 7 4 2

16
题目意思是：求出循环n次中最小的逆序数。
#include<stdio.h>
#define N 5000
struct nod
{
    int fmax,fmin,rmax,rmin,w;//分别代表序列的前方比w大的个数，小的个数，后方比w大的个数，小的个数
}node[N+5];
int tree[4*N],fmax,fmin;
void biulde(int l,int r,int k)
{
    int m=(l+r)/2;
    tree[k]=0;
    if(l==r) return ;
    biulde(l,m,k*2); biulde(m+1,r,k*2+1);
}
void updata(int l,int r,int k,int w)
{
    int m=(l+r)/2;
    tree[k]++;
    if(l==r)
        return ;
    if(w<=m)//往左走
    {
        fmax+=tree[k*2+1];//记录比w大的个数
        updata(l,m,k*2,w);
    }
    if(w>m)//往右走
    {
        fmin+=tree[k*2];//记录比w小的个数
        updata(m+1,r,k*2+1,w);
    }
}
int main()
{
    int n,minI,I;
    while(scanf("%d",&n)>0)
    {
        biulde(1,n,1); I=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&node[i].w); node[i].w++;
            fmax=0; fmin=0;
            updata(1,n,1,node[i].w);
            node[i].fmax=fmax;//前面输入的比w大的个数
            node[i].fmin=fmin;//前面输入的比w小的个数
            node[i].rmax=(n-node[i].w)-fmax;//根据前面的可推出后面比w大的个数
            node[i].rmin=(node[i].w-1)-fmin;//根据前面的可推出后面比w小的个数
            I+=fmax;//记录输入序列的总逆序数
        }
        minI=I;//记录n个（循环）序列的最小的总逆序数
        for(int i=1;i<n;i++)//把第i个数放到序列的最后
        {
            I=I-node[i].rmin;//移动后，对后方小于w(i)的逆序数的影响
            I=I-node[i].fmin;//移动后，对己经循环的前方数(小于w(i))的逆序数的影响
            I=I+node[i].fmax;//移动后，对本身(小于己经循环的前方数的逆序数)的影响
            I=I+node[i].rmax;//移动后，对本身(小于还没循环数的逆序数)的影响
            if(I<minI) minI=I;
        }
       printf("%d\n",minI);
    }
}