HDUOJ---1754 Minimum Inversion Number （单点更新之求逆序数）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9342    Accepted Submission(s): 5739

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

10

1 3 6 9 0 8 5 7 4 2

 

 

Sample Output

16

 

 

Author

CHEN, Gaoli

 

 

Source

ZOJ Monthly, January 2003

 

代码：

 1 //线段树实现单点更新，并求和

 2 #include<stdio.h>

 3 #define maxn 5001

 4 struct node{

 5   int lef,rig,sum;

 6   int mid(){ return lef+((rig-lef)>>1) ;}

 7 };

 8  node seg[maxn<<2];

 9  int aa[maxn+3];

10 void build(int left,int right,int p )

11 {

12     seg[p].lef=left;

13     seg[p].rig=right;

14     seg[p].sum=0;

15     if(left==right) return ;

16     int mid=seg[p].mid();

17     build(left,mid,p<<1);

18     build(mid+1,right,p<<1|1);

19 }

20 void updata(int pos,int p,int val)

21 {

22     if(seg[p].lef==seg[p].rig)

23     {

24       seg[p].sum+=val;

25       return ;

26     }

27     int mid=seg[p].mid();

28     if(pos<=mid) updata(pos,p<<1,val);

29     else updata(pos,p<<1|1,val);

30    seg[p].sum=seg[p<<1].sum+seg[p<<1|1].sum;

31 }

32 int query(int be ,int en,int p)

33 {

34     if(be<=seg[p].lef&&seg[p].rig<=en)

35         return seg[p].sum;

36     int mid=seg[p].mid();

37     int res=0;

38     if(be<=mid) res+=query(be ,en ,p<<1);

39     if(mid<en)  res+=query(be ,en ,p<<1|1);

40     return res;

41 }

42 int main()

43 {

44     int nn,i,ans;

45     while(scanf("%d",&nn)!=EOF)

46     {

47       ans=0;

48       build(0,nn-1,1);

49      for(i=1;i<=nn;i++)

50      {

51        scanf("%d",&aa[i]);

52        updata(aa[i],1,1);

53        if(aa[i]!=nn-1) ans+=query(aa[i]+1,nn-1,1); //统计比其大的数

54      }

55      int min=ans;

56      for(i=1;i<=nn;i++)

57      {

58        ans+=nn-2*aa[i]-1;

59        if(min>ans) min=ans;

60      }

61      printf("%d\n",min);

62     }

63     return 0 ;

64 }

View Code