hdu 1394 Minimum Inversion Number（樹狀數組）

                                        Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6937    Accepted Submission(s): 4233

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  

Recommend Ignatius.L

題意：将一個n個元素的排列，可以将第一個元素放到最後一個元素後面得出n個排列，求這n個排列的最少逆序數對 題解：用樹狀數組，按順序插入，求出排列的逆序數對temp，模拟将第一個元素移走，那麼逆序數對就減少sum（a-1）個，将這個數放到最後一位，逆序數對就增加（n-1）-sum（a-1）對，是以每次轉變就有temp=temp+n-1-2*sum（a【i】-1），依次可以算出最少的逆序數對值

#include<stdio.h>
#include<string.h>
#define MAX 1000005
int tree[MAX],a[5005];
int low_bit(int x)
{
    return x&(-x);
}
void add(int x,int y)
{
    while(x<MAX)
    {
        tree[x]+=y;
        x+=low_bit(x);
    }
}
int ques(int x)
{
    int temp=0;
    while(x>0)
    {
        temp+=tree[x];
        x-=low_bit(x);
    }
    return temp;
}
int main()
{
    int temp,sum,n,i;

    while(scanf("%d",&n)>0)
    {
        memset(tree,0,sizeof(tree));
        for(i=1;i<=n;i++) scanf("%d",a+i);
        for(temp=0,i=n;i>=1;i--)
        {
            a[i]++;
            temp=temp+ques(a[i]-1);
            add(a[i],1);
        }
        sum=temp;
        for(i=1;i<n;i++)
        {
            temp=temp+n-1-2*ques(a[i]-1);
            if(temp<sum) sum=temp;
        }
        printf("%d\n",sum);
    }

    return 0;
}