辣么这个题可是真难

首先利用Hash表判断是否可以用

T2则是利用并查集的siz

好难啊

枚举区间段可以用BIT

我实际就没懂

#include<bits/stdc++.h>
using namespace std;
typedef int INT;
#define int long long
inline void read(int &x){x=0;int f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}x*=f;
}
const int N=1e6+1000;
struct BIT{int T[N];inline int lowbit(int x){return x&(-x);}inline void update(int x,int val){while(x<N){T[x]+=val;x+=lowbit(x);}}inline int query(int x){if(x==-1) return 0;int ret=0;while(x){ret+=T[x];x-=lowbit(x);}return ret;}
}T;
struct Heap{priority_queue<int>In,Del;void Insert(int x){In.push(x);}void Delete(int x){Del.push(x);}int Top(){while(Del.size()&&In.top()==Del.top()){In.pop();Del.pop();};return (In.size()?In.top():0);}
}q;int pre[N]={};
int v[N]={};
int fa[N]={};
int n,m;
vector<int> A[N];
int pos[N]={};
int siz[N]={};
int find(int x){if(fa[x]==x)return x;else return fa[x]=find(fa[x]);
}
INT main(){read(n);read(m);for(int i=1;i<=n;i++){read(v[i]);fa[i]=i;A[v[i]].push_back(i);pos[i]=A[v[i]].size()-1;siz[i]=1;}fa[0]=1;fa[n+1]=n+1;siz[n+1]=1;int Up_Low=n;int sum=0;
//  cout<<find(1);/*for(int i=1;i<n;i++){if(pos[i])  q.Delete(A[v[i]][pos[i]-1]);if(pos[i]==(int)A[v[i]].size()-1){for(int j=find(A[v[i]][0]);j<i;j=fa[j])   T.update(j,1),siz[find(j+1)]+=siz[j],fa[j]=fa[j+1];}else    q.Insert(i);int k=q.Top();sum+=i-k-(T.query(i-1)-T.query(k-1));cout<<sum<<'\n';
//        j=find(max(k,i-n/2));
//        if(j<i)  ans=min(ans,abs(n-2*(i-j)));
//        j-=siz[j];
//        if(j>=k) ans=min(ans,abs(n-2*(i-j)));}*/for(int i=1;i<n;i++){
//      cout<<i<<'\n';if(pos[i])q.Delete(A[v[i]][pos[i]-1]);
//      cout<<"234789478"<<'\n';
//      cout<<A[v[i]].size()<<'\n';
//      if(pos[i]==(int)A[v[i]].size()-1){
//          cout<<"in"<<'\n';for(int j=find(A[v[i]][0]);j<i;j=fa[j]){
//              cout<<j<<" k"<<'\n';T.update(j,1);siz[find(j+1)]+=siz[j];fa[j]=fa[j+1];
//              cout<<"j= "<<j<<'\n';}}else q.Insert(i);int now=q.Top();
//      cout<<"13"<<'\n';sum+=i-now-(T.query(i-1)-T.query(now-1));
//      cout<<sum<<'\n';int idx=find(max(now,i-n/2));if(idx<i) Up_Low=min(Up_Low,abs(n-2*(i-idx)));idx-=siz[idx];if(idx>=now) Up_Low=min(Up_Low,abs(n-2*(i-idx)));}cout<<sum<<" "<<Up_Low;
}