CF&&CC百套计划3 Codeforces Round #204 (Div. 1) D. Jeff and Removing Periods

#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>

using namespace std;

#define N 100002

int val[N],last[N];
int pre[N],fl[N];
int bac[N],fr[N];

struct node
{
    int l,r;
    int id;
}e[N];

int bl[N];

int sum[N];

int ans[N];

void read(int &x)
{
    x=0; char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) { x=x*10+c-‘0‘; c=getchar(); }
}

bool cmp(node p,node q)
{
    if(bl[p.l]!=bl[q.l]) return bl[p.l]<bl[q.l];
    return p.r<q.r; 
}

int main()
{
    int n;
    read(n);
    int x,y;
    for(int i=1;i<=n;++i)
    {
        read(val[i]);
        pre[i]=last[val[i]];
        last[val[i]]=i;
        if(pre[pre[i]]-pre[i]==pre[i]-i || !pre[i])    fl[i]=fl[pre[i]];
        else fl[i]=pre[pre[i]];
    }    
    for(int i=1;i<=n;++i) last[i]=n+1;
    fr[n+1]=n+1;
    for(int i=n;i;--i)
    {
        bac[i]=last[val[i]];
        last[val[i]]=i;
        if(bac[bac[i]]-bac[i]==bac[i]-i || bac[i]==n+1) fr[i]=fr[bac[i]];
        else fr[i]=bac[bac[i]];
    }
    int m;
    read(m);
    for(int i=1;i<=m;++i)
    {
        read(e[i].l);
        read(e[i].r);
        e[i].id=i;
    }
    int siz=sqrt(n);
    for(int i=1;i<=n;++i) bl[i]=(i-1)/siz+1;
    sort(e+1,e+m+1,cmp);
    int L=1,R=0,tmp=0,dengcha=0;
    for(int j=1;j<=m;++j)
    {
        x=e[j].l; y=e[j].r;
    
            for(int i=L-1;i>=x;--i)
            {
                if(!sum[val[i]])
                {
                    dengcha++;
                    tmp++;
                }
                else
                {
                    if(fr[i]<=R && fr[bac[i]]>R) dengcha--;
                }
                sum[val[i]]++;
            }
            for(int i=L;i<x;++i)
            {
                if(sum[val[i]]==1)
                {
                    dengcha--;
                    tmp--;
                }
                else
                {
                    if(fr[i]<=R && fr[bac[i]]>R) dengcha++;
                }
                sum[val[i]]--; 
            }
            for(int i=R;i>y;--i)
            {
                if(sum[val[i]]==1)
                {
                    dengcha--;
                    tmp--;
                }
                else
                {
                    if(fl[i]>=x && fl[pre[i]]<x) dengcha++;
                }
                sum[val[i]]--;
            }
            for(int i=R+1;i<=y;++i)
            {
                if(!sum[val[i]])
                {
                    dengcha++;
                    tmp++;
                }
                else
                {
                    if(fl[i]>=x && fl[pre[i]]<x) dengcha--;
                }
                sum[val[i]]++;
            }
        L=x; R=y;
        ans[e[j].id]=tmp+1;
        if(dengcha>0) ans[e[j].id]--;
    }
    for(int i=1;i<=m;++i) cout<<ans[i]<<‘\n‘;
}

Cosider a sequence, consisting of n integers: a1, a2, ..., an. Jeff can perform the following operation on sequence a:

• take three integers v, t, k (1 ≤ v, t ≤ n; 0 ≤ k; v + tk ≤ n), such that av = av + t, av + t = av + 2t, ..., av + t(k - 1) = av + tk;
• remove elements av, av + t, ..., av + t·k from the sequence a, the remaining elements should be reindexed a1, a2, ..., an - k - 1.
• permute in some order the remaining elements of sequence a.

A beauty of a sequence a is the minimum number of operations that is needed to delete all elements from sequence a.

Jeff‘s written down a sequence of m integers b1, b2, ..., bm. Now he wants to ask q questions. Each question can be described with two integers li, ri. The answer to the question is the beauty of sequence bli, bli + 1, ..., bri. You are given the sequence b and all questions. Help Jeff, answer all his questions.