C - Minimum Inversion Number

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

先求出逆序数，在一个个dp

 #include"iostream"
 #include"cstdio"
 #include"string"
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define MAX 5004
 int s[MAX];
 int sum;
 struct node
 {
     int l;
     int r;
     int va;
 }tree[MAX<<];
 void build(int l,int r,int rt)
 {
     tree[rt].l=l;
     tree[rt].r=r;
     tree[rt].va=;
     if(tree[rt].l==tree[rt].r)
        return ;
     int mid=(tree[rt].l+tree[rt].r)>>;
     build(lson);
     build(rson);
 }
 void updata(int x,int rt)
 {
     if(tree[rt].l==tree[rt].r)
     {
         tree[rt].va=;
         return;
     }
     int mid=(tree[rt].l+tree[rt].r)>>;
     if(x<=mid)
       updata(x,rt<<);
     else
        updata(x,rt<<|);
     tree[rt].va=tree[rt<<].va+tree[rt<<|].va;
 }
 void query(int l,int r,int rt)
 {
     if(tree[rt].l==l&&tree[rt].r==r)
     {
         sum+=tree[rt].va;
         return;
     }
     int mid=(tree[rt].l+tree[rt].r)>>;
     if(r<mid)
       query(l,r,rt<<);
     else if(l>mid)
        query(l,r,rt<<|);
     else
     {
         query(lson);
         query(rson);
     }
 }
 int main()
 {
     int n,i,t,m;
     while(scanf("%d",&n)!=EOF)
     {
         build(,n-,);
         sum=;
         for(i=;i<n;i++)
         {
             scanf("%d",&s[i]);
             query(s[i],n-,);
             updata(s[i],);
         }
         m=sum;
         for(i=;i<n;i++)
         {
             sum=sum-s[i]+n--s[i];
             if(m>sum)
               m=sum;
         }
         printf("%d\n",m);
     }
     return ;
 }