hdu 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5634    Accepted Submission(s): 3449

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

分析：求逆序数问题，先求出原始序列的逆序数

暴力n^2

#include<cstdio>
int a[5000];
int main()
{
	int n,i,j,now,ans;
	while(~scanf("%d",&n)){
		for(i=0;i<n;i++)scanf("%d",a+i);
		now=0;
		for(i=1;i<n;i++)for(j=0;j<i;j++)if(a[j]>a[i])now++;
		ans=now;
		for(i=1;i<n;i++){
			now=now+n-1-a[i-1]*2;
			if(now<ans)ans=now;
		}
		printf("%d\n",ans);
	}
	return 0;
}

归并排序nlogn

#include<cstdio>
const int N=50001;
int a[N],tmp[N],b[N];
int co;
void merge(int first,int mid,int last)
{
	int i=first,j=mid+1,cur=1;
	while(i<=mid&&j<=last){
		if(a[i]<a[j])tmp[cur++]=a[i++];
		else{
			tmp[cur++]=a[j++];
			co+=mid-i+1;
		}
	}
	while(i<=mid)tmp[cur++]=a[i++];
	while(j<=last)tmp[cur++]=a[j++];
	for(i=first,j=1;i<=last;i++,j++)a[i]=tmp[j];
}
void mergesort(int first,int last)
{
	if(first<last){
		int mid=(first+last)>>1;
		mergesort(first,mid);
		mergesort(mid+1,last);
		merge(first,mid,last);
	}
}
int main()
{
	int n,ans,t,i;
	while(~scanf("%d",&n)){
		for(i=0;i<n;i++)scanf("%d",a+i),b[i]=a[i];
		co=0;
		mergesort(0,n-1);
		t=ans=co;
		for(i=1;i<n;i++){
			t=t+n-1-b[i-1]*2;
			if(t<ans)ans=t;
		}
		printf("%d\n",ans);
	}
	return 0;
}

树状数组

#include<cstdio>
#include<cstring>
int a[5001],b[5001],n;
int lowbit(int x){return x&(-x);}
void update(int x,int v){
	while(x<=n){
		a[x]+=v;
		x+=lowbit(x);
	}
}
int sum(int x){
	int s=0;
	while(x>0){
		s+=a[x];
		x-=lowbit(x);
	}
	return s;
}
int main()
{
	int i,x,ans,t;
	while(~scanf("%d",&n)){
		memset(a,0,sizeof(a));ans=0;
		for(i=1;i<=n;i++)scanf("%d",&b[i]),update(b[i]+1,1),ans+=(sum(n)-sum(b[i]+1));
		t=ans;
		for(i=1;i<n;i++){
			t=t+n-1-b[i]*2;
			if(t<ans)ans=t;
		}
		printf("%d\n",ans);
	}
	return 0;
}

线段树

#include<cstdio>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int N=5001;
int a[N<<2],b[N];
void pushup(int rt){
	a[rt]=a[rt<<1]+a[rt<<1|1];
}
void build(int l,int r,int rt){
	a[rt]=0;
	if(l==r)return;
	int m=(l+r)>>1;
	build(lson);
	build(rson);
}
void update(int p,int l,int r,int rt){
	if(l==r){a[rt]++;return;}
	int m=(l+r)>>1;
	if(p<=m)update(p,lson);
	else update(p,rson);
	pushup(rt);
}
int query(int L,int R,int l,int r,int rt){
	if(L<=l&&r<=R)return a[rt];
	int m=(l+r)>>1,ret=0;
	if(L<=m)ret+=query(L,R,lson);
	if(R>m)ret+=query(L,R,rson);
	return ret;
}
int main()
{
	int n,i,t,ans;
	while(~scanf("%d",&n)){
		build(1,n,1);
		ans=0;
		for(i=0;i<n;i++){
			scanf("%d",b+i);
			ans+=query(b[i]+1,n,1,n,1);
			update(b[i]+1,1,n,1);
		}
		t=ans;
		for(i=0;i<n-1;i++){
			t+=n-1-b[i]*2;
			if(t<ans)ans=t;
		}
		printf("%d\n",ans);
	}
	return 0;
}