HDU 1394 Minimum Inversion Number

 Minimum Inversion Number

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1394

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

<pre name="code" class="cpp">#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define MS(x,y)memset(x,y,sizeof(x))#define max(a,b) a>b?a:b#define min(a,b) a<b?a:b#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long LL;const int INF=0x7ffffff;const int MX = 5000+10;int sum[MX<<2];void PushUp(int rt) {    sum[rt]=sum[rt<<1]+sum[rt<<1|1];  }void Build(int l,int r,int rt) {    sum[rt]=0;                if(r==l) return ;    int    m=(r+l)>>1;    Build(lson);    Build(rson);}void UpData(int p,int l,int r,int rt) {    if(r==l) {                    sum[rt]++;        return ;    }    int m=(r+l)>>1;    if(p<=m) UpData(p,lson);      if(p >m) UpData(p,rson);    PushUp(rt);}int Query(int L,int R,int l,int r,int rt) {    if(L<=l&&R>=r)                return sum[rt];    int m=(r+l)>>1;    int ret=0;    if(L<= m) ret += Query(L,R,lson);      if(R > m) ret += Query(L,R,rson);      return ret;}int x[MX];int main() {    int n;    int sums;    char s[2];    while(~scanf("%d",&n)) {        sums=0;                    Build(0,n-1,1);                for(int i=0; i<n; i++) {            scanf("%d",&x[i]);            sums+=Query(x[i],n-1,0,n-1,1);            UpData(x[i],0,n-1,1);        }        int ret=sums;        for(int i=0; i<n; i++) {            sums=sums+n-2*x[i]-1;            ret=min(ret,sums);        }        printf("%d\n",ret);    }    return 0;}