hdu 1394（求逆序数）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16401    Accepted Submission(s): 9975

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

 

#include<stdio.h>#include<string.h>struct tree{    int l,r,n;} s[5000010];void in(int i,int l,int r){    s[i].l=l;    s[i].r=r;    s[i].n=0;    if(l==r)        return ;    int mid=(l+r)/2;    in(i*2,l,mid);    in(i*2+1,mid+1,r);}void _insert(int i,int m){    if(s[i].l==m&&s[i].r==m)    {        s[i].n=1;        return ;    }    int mid=(s[i].l+s[i].r)/2;    if(m<=mid)    {        _insert(i*2,m);    }    else    {        _insert(i*2+1,m);    }    s[i].n=s[i*2].n+s[i*2+1].n;}int _add(int i,int l,int r){    if(s[i].l==l&&s[i].r==r)    {        return s[i].n;    }    int mid=(s[i].l+s[i].r)/2;    if(l<=mid)    {        if(r<=mid)            return _add(i*2,l,r);        else            return _add(i*2,l,mid)+_add(i*2+1,mid+1,r);    }    else        return _add(i*2+1,l,r);}int main(){    int n,a[5010];    while(~scanf("%d",&n))    {        in(1,0,n-1);        int sum=0;        for(int i=0; i<n; i++)        {            scanf("%d",&a[i]);            if(a[i]+1<n)            {                sum+=_add(1,a[i]+1,n-1);            }            _insert(1,a[i]);        }        int min=sum;        for(int i=0; i<n; i++)        {            sum=sum+n-1-2*a[i];  //当a[i]由第一个变为最后一个时，要加上a[i]后面大于a[i]的数的个数，有n-1-a[i]个，要            if(min>sum)              //减去a[i]后面小于a[i]的数的个数，有a[i]个（注意i是从0开始的）                min=sum;        }        printf("%d\n",min);    }    return 0;}