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.
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 }