Luogu P5048 [Ynoi2019模拟赛]Yuno loves sqrt technology III 分块

这才是真正的$Nsqrt{N}$吧$qwq$

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记录每个数$vl$出现的位置$s[vl]$，和每个数$a[i]=vl$是第几个$vl$,记为$P[i]$，然后预处理出块$[i,j]$区间的答案$f[i][j]$;

对于$[l,r]$,现将$ans$设为$[l,r]$中整块的答案；对于散块，将散块中的每个数$a[i]=vl$,有对应的$P[i]$，我们用$s[vl]$检查第$P[i]+ans+1$个$vl$对应的位置是否在$[l,r]$，如果在的话就更新答案$ans=ans+1$，并重复上述过程，直到$P[i]+ans+1$不在$[l,r]$中。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<queue>
#include<map>
#include<set>
#define ull unsigned long long
#define ll long long
#define R register int
#define pause (for(R i=1;i<=10000000000;++i))
#define OUT freopen("out.out","w",stdout);
using namespace std;
namespace Fread {
    static char B[1<<15],*S=B,*D=B;
    #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
    inline int g() {
        R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
        do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
    } inline bool isempty(const char& ch) {return ch<=36||ch>=127;}
    inline void gs(char* s) {register char ch; while(isempty(ch=getchar())); do *s++=ch; while(!isempty(ch=getchar()));}
}using Fread::g; using Fread::gs;
const int N=500010,M=710;
int n,m,T,tot,lst; int cnt[N],pos[N],P[N],a[N],b[N],f[M][M];
vector <int> s[N];
#define pb push_back
signed main() {
    n=g(),m=g(),T=pow(n,1/2.0); for(R i=1;i<=n;++i) a[i]=g(); memcpy(b,a,sizeof(int)*(n+1));
    sort(b+1,b+n+1); tot=unique(b+1,b+n+1)-b-1; 
    for(R i=1;i<=n;++i) a[i]=lower_bound(b+1,b+tot+1,a[i])-b;
    for(R i=1;i<=n;++i) s[a[i]].pb(i);
    for(R i=1;i<=n;++i) P[i]=++cnt[a[i]]-1; memset(cnt,0,sizeof(cnt));
    for(R i=1;i<=n;++i) pos[i]=(i-1)/T+1;
    for(R p=1,L=pos[n],mx=0;p<=L;++p,mx=0,memset(cnt,0,sizeof(cnt))) for(R t=p;t<=L;++t) {
        for(R i=(t-1)*T+1,LL=min(t*T,n);i<=LL;++i) mx=max(++cnt[a[i]],mx);
        f[p][t]=mx;
    } for(R i=1,l,r;i<=m;++i) { R ans=0;
        l=g()^lst,r=g()^lst;
        if(pos[l]==pos[r]) {
            for(R i=l;i<=r;++i) ans=max(ans,++cnt[a[i]]); for(R i=l;i<=r;++i) cnt[a[i]]=0;
        } else { R p=pos[l]+1,q=pos[r]-1; ans=f[p][q];
            for(R i=l,L=pos[l]*T;i<=L;++i) { R tmp=P[i];
                while((tmp+ans)<s[a[i]].size()&&s[a[i]][tmp+ans]<=r) ++ans;
            } for(R i=q*T+1;i<=r;++i) { R tmp=P[i];
                while((tmp-ans)>=0&&s[a[i]][tmp-ans]>=l) ++ans;
            }
        } printf("%d
",lst=ans); 
    }
}

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2019.07.03