HDU1394 逆序数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19603    Accepted Submission(s): 11774

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

 题意：

给长度是n的序列，每次将第一个数放到最后，直到数列再次变成最初的数列，求过程中最小的逆序数。

代码：

//求出最初的逆序数sum，每次操作将a[i]放到最后时更新sum：
//加上a[i]向后的顺序数（比他大的数的个数），减去a[i]向后的
//逆序数（比她小的数的个数），加上a[i]向前的（他前面的数放到后面了）
//顺序数减去a[i]向前的逆序数。所以两遍树状数组就解决了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=5000;
int n,c[maxn*4+10],a[maxn+10],sh[maxn+10],ni[maxn+10];
int lowbit(int x) {return x&(-x);}
void add(int id)
{
    while(id<=5000){
        c[id]++;
        id+=lowbit(id);
    }
}
int query(int id)
{
    int s=0;
    while(id>0){
        s+=c[id];
        id-=lowbit(id);
    }
    return s;
}
int main()
{
    int n;
    while(scanf("%d",&n)==1){
        int sum=0;
        memset(c,0,sizeof(c));
        memset(sh,0,sizeof(sh));
        memset(ni,0,sizeof(ni));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            a[i]++;
            add(a[i]);
            sh[i]=query(a[i]-1);
            ni[i]=i-sh[i]-1;
            sum+=ni[i];
        }
        //for(int i=1;i<=n;i++) cout<<sh[i]<<" "<<ni[i]<<endl;
        memset(c,0,sizeof(c));
        for(int i=n;i>=1;i--){
            add(a[i]);
            int tmp=query(a[i]-1);
            sh[i]+=tmp;
            ni[i]+=n-i+1-tmp-1;
        }
        //for(int i=1;i<=n;i++) cout<<sh[i]<<" "<<ni[i]<<endl;
        int ans=sum;
        for(int i=1;i<=n;i++){
            sum=sum+ni[i]-sh[i];
            ans=min(ans,sum);
        }
        printf("%d
",ans);
    }
    return 0;
}