hdu1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8498    Accepted Submission(s): 5237

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

 

线段树求逆序数。基础线段树

#include<cstdio>#include<iostream>#include<cstring>using namespace std;#define inf 0x3f3f3f3fint a[55555],sum[55555*3],h[55555];int n,tmp,ans,tot;void update(int o,int x,int y,int l,int r){    if (l<=x&&y<=r)    {        sum[o]++;        return;    }    int mid=(x+y)>>1;    if (l<=mid) update(o*2,x,mid,l,r);    if (mid<r ) update(o*2+1,mid+1,y,l,r);    sum[o]=sum[o*2]+sum[o*2+1];}void query(int o,int x,int y,int l,int r){    if (l<=x&&y<=r)    {        tmp+=sum[o];        return;    }    int mid=(x+y)>>1;    if (l<=mid)  query(o*2,x,mid,l,r);    if (mid< r)  query(o*2+1,mid+1,y,l,r);}int main(){    while (scanf("%d",&n)!=EOF)    {        for (int i=1; i<=n; i++)            scanf("%d",&a[i]);        memset(sum,0,sizeof(sum));        tot=0;        for (int i=1; i<=n; i++)        {            tmp=0;            query(1,0,n,a[i]+1,n);            tot+=tmp;            update(1,0,n,a[i],a[i]);        }        ans=tot;        for (int i=1; i<n; i++)        {            tot=tot-a[i]+(n-1-a[i]);            ans=min(ans,tot );        }        printf("%d\n",ans);    }    return 0;}