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

时间:2022-05-05
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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("%dn",min);
62     }
63     return 0 ;
64 }