showproblem.php?pid=1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8205 Accepted Submission(s): 5041

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

1 3 6 9 0 8 5 7 4 2

Sample Output

#include<stdio.h>

#define maxn 500004

struct node

{

    int left,right;

    int sum;

};

int val[maxn];

int n;

node tree[*maxn];

void build(int left,int right,int i)

{

    tree[i].left =left;

    tree[i].right =right;

    tree[i].sum =;

    if(tree[i].left ==tree[i].right )

        return ;

    build(left,(left+right)/,*i);

    build((left+right)/+,right,*i+);

}

void update(int r,int j)

{

    tree[r].sum++;

    if(tree[r].left==tree[r].right)

        return ;

int mid=(tree[r].left +tree[r].right )/;

      if(j<=mid)

          update(r*,j);

      if(j>mid)

          update(r*+,j);

}

int getsum(int p, int left, int right)

{

    if (left > right)

        return ;

    if (tree[p].left == left && tree[p].right == right)

        return tree[p].sum;

    int m = (tree[p].left + tree[p].right) / ;

    if (right <= m)

        return getsum(p*, left, right);

    else if (left > m)

        return getsum(p*+, left, right);

    else return getsum(p*, left, m) + getsum(p*+, m + , right);

}

int main()

{

    while (~scanf("%d",&n))

    {

        build(,n - ,);

        int i,sum =,ans;

        for (i = ;i<= n; i++)

        {

            scanf("%d", &val[i]);

            sum += getsum(, val[i], n - );

            update(, val[i]);

        }

        ans = sum;

        for (i = ; i <= n; i++)

        {

            sum = sum + (n - val[i] - ) - val[i];

            ans = sum < ans ? sum : ans;

        }

        printf("%d\n", ans);

    }

    return ;

}

巴特西

HDU1394-Minimum Inversion Number

http://acm.hdu.edu.cn/showproblem.php?pid=1394

Minimum Inversion Number

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