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
求逆序数的最小值
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 using namespace std;
5 const int maxn=5010;
6 int sum[maxn<<2];
7 void push_up(int rt)
8 {
9 sum[rt]=sum[rt<<1]+sum[rt<<1 | 1];
10 }
11 void build(int rt,int first,int end)
12 {
13 sum[rt]=0;
14 if(first==end)
15 return ;
16 int mid=(first+end)>>1;
17 build(rt<<1,first,mid);
18 build(rt<<1 | 1,mid+1,end);
19 }
20 int quary(int rt,int first,int end,int left,int right)
21 {
22 if(left<=first && right>=end)
23 return sum[rt];
24 int mid=(first+end)>>1;
25 int ans=0;
26 if(left<=mid)
27 ans+=quary(rt<<1,first,mid,left,right);
28 if(right>mid)
29 ans+=quary(rt<<1 | 1,mid+1,end,left,right);
30 return ans;
31 }
32 void update(int rt,int first,int end,int a)
33 {
34 if(first==end)
35 {
36 sum[rt]++;
37 return ;
38 }
39 int mid=(first+end)>>1;
40 if(a<=mid)
41 update(rt<<1,first,mid,a);
42 else
43 update(rt<<1 | 1,mid+1,end,a);
44 push_up(rt);
45 }
46 int a[maxn];
47 int main()
48 {
49 int n;
50 while(cin>>n)
51 {
52 build(1,0,n-1);
53 int sum=0;
54 for(int i=0;i<n;i++)
55 {
56 cin>>a[i];
57 sum+=quary(1,0,n-1,a[i],n-1);
58 update(1,0,n-1,a[i]);
59 }
60 int ans=sum;
61 for(int i=0;i<n;i++)
62 {
63 sum+=n-a[i]-a[i]-1;
64 ans=min(ans,sum);
65 }
66 cout<<ans<<endl;
67 }
68 return 0;
69 }