hdu 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16798    Accepted Submission(s): 10215

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 <iostream>#include <cstdio>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int N = 5555;int sum[N<<2], x[N];void build(int l, int r, int rt);void pushup(int rt);void update(int p,int l, int r,int rt);int query(int L,int R,int l, int r,int rt);int main(){    int n;    while(~scanf("%d",&n))    {        build(0,n-1,1);        int s=0;        for(int i=0;i<n;i++)        {            scanf("%d",&x[i]);            s+=query(x[i],n-1,0,n-1,1);            update(x[i],0,n-1,1);        }        int r=s;        for(int i=0;i<n;i++)        {            s+=n-x[i]-1-x[i];            r=min(r,s);        }        cout<<r<<endl;    }    return 0;}void build(int l, int r, int rt){    sum[rt]=0;    if(l==r)    {        return ;    }    int m=(l+r)/2;    build(lson);    build(rson);    return ;}void pushup(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];    return ;}int query(int L,int R,int l, int r,int rt){    if(L<=l&&r<=R)    {        return sum[rt];    }    int m=(l+r)/2;    int s=0;    if(L<=m)    {        s+=query(L,R,lson);    }    if(R>m)    {        s+=query(L,R,rson);    }    return s;}void update(int p,int l, int r,int rt){    if(l==r)    {        sum[rt]++;        return ;    }    int m=(l+r)/2;    if(p<=m)    {        update(p,lson);    }    else    {        update(p,rson);    }    pushup(rt);    return ;}