Minimum Inversion Number

線段樹解決逆序數問題

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

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=5005;
struct node{
	int sum,l,r;
}tr[maxn<<2];
int x[maxn];
void pushup(int m){
	tr[
 
m].sum=tr[m<<1].sum+tr[m<<1|1].sum;
}
void build(int m,int l,int r){
	tr[m].l=l;
	tr[m].r=r;
	if(l==r){
		tr[m].sum=0;
		return ;
	}
	int mid=(l+r)>>1;
	build(m<<1,l,mid);
	build(m<<1|1,mid+1,r);
	pushup(m);
}
void updata(int m,int inx,int val){
	if(tr[m].l==inx&&tr[m].r==inx){
		tr[m].sum+=val;
		return ;
	}
	int mid=(tr[m].l+tr[m].r)>>1;
	if(inx<=mid) updata(m<<1,inx,val);
	else updata(m<<1|1,inx,val);
	pushup(m);
}
int query(int m,int l,int r){
	if(tr[m].l==l&&tr[m].r==r){
		return tr[m].sum;
	}
	int mid=(tr[m].l+tr[m].r)>>1;
	int temp;
	if(r<=mid) temp=query(m<<1,l,r);
	else if(l>mid) temp=query(m<<1|1,l,r);
	else temp=query(m<<1,l,mid)+query(m<<1|1,mid+1,r);
	return temp;
}
int main(){
	int n;
	while(~scanf("%d",&n)){
		build(1,0,n);
		int sum=0;
		for(int i=0;i<n;i++){
			scanf("%d",&x[i]);
		    sum+=query(1,x[i]+1,n);
			updata(1,x[i],1);	
		}	
		int min=sum;
		for(int i=0;i<n-1;i++){
			sum+=n-1-x[i]-x[i];
			if(sum<min) min=sum;
		}
		printf("%d\n",min);
	}
	return 0;
}