hdu 1394 Minimum Inversion Number 最小逆序数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15355    Accepted Submission(s): 9366

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

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

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#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int  maxn=5000;int n;int a[maxn+20];int sum[4*maxn+20];int tot;void update(int num,int le,int ri,int x){if(le==ri){sum[num]++;return;}int mid=(le+ri)>>1;if(x<=mid)   update(num*2,le,mid,x);else   update(num*2+1,mid+1,ri,x);sum[num]=sum[num<<1]+sum[(num<<1)+1];}int query(int num,int le,int ri,int x,int y){if(x<=le&&ri<=y){return sum[num];}   int ret=0;   int mid=(le+ri)>>1;   if(x<=mid)   ret+=query(num*2  ,le   ,mid,x,y);   if(y> mid)   ret+=query(num*2+1,mid+1,ri ,x,y);   return ret;}int main(){    while(~scanf("%d",&n)){memset(sum,0,sizeof sum);tot=0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);tot+=query (1,0,maxn,a[i]+1,maxn);     update(1,0,maxn,a[i]);}int mini=tot;for(int i=1;i<=n-1;i++){int sm=query(1,0,maxn,0,a[i])-1;int la=n-1-sm;tot=tot-sm+la;mini=min(mini,tot);}printf("%d\n",mini);}    return 0;}