AT1219 歴史の研究 [回滚莫队]

$歴史の研究$

题目描述见链接 .

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$color{red}{正解部分}$

直接使用 莫队, 但是这道题目删除时不好寻找次大值, 考虑 回滚莫队,

回滚莫队 的处理方式与 莫队 大致是相同的, 具体来说

对两个相邻的询问区间 $[l',r'], [l, r]$,

1. 若 $bk\_id[l] = bk\_id[r]$, 直接 $O(sqrt{N})$ 暴力求解 .
2. 若 $bk\_id[l'] ot= bk\_id[l]$, 暴力将左端点和右端点移动到 $l+1, l$ .
3. 若 $bk\_id[l'] = bk\_id[l]$, 暴力处理左端点对答案的贡献, 处理完将左端点归位,
   由于右端点是有序的, 可以直接将右端点从 $r'$ 移动到 $r$, 无需归位,
   每个块移动右端点 $O(N)$ 次, 复杂度 $O(Nsqrt{N})$ .

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$color{red}{实现部分}$

#include<bits/stdc++.h>
#define reg register
typedef long long ll;

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 1e5 + 5;

int N;
int Q_;
int Len;
int bkcnt;
int A[maxn];
int B[maxn];
int llim[maxn];
int rlim[maxn];
int bkid[maxn];
int Tong[maxn];
int Tong2[maxn];

ll Ans;
ll res[maxn];

struct Que{ int l, r, id; } que[maxn];

bool cmp(Que a, Que b){ return bkid[a.l]==bkid[b.l]?a.r<b.r:bkid[a.l]<bkid[b.l]; }

void Add(const int &x){ Ans = std::max(Ans, 1ll*B[A[x]]*(++ Tong[A[x]])); }

void Del(const int &x){ Tong[A[x]] --; }

void write(ll x){ if(x > 9) write(x/10); putchar(x%10 + '0'); }

int main(){
        N = read(), Q_ = read();
        for(reg int i = 1; i <= N; i ++) B[i] = A[i] = read();
        std::sort(B+1, B+N+1); Len = std::unique(B+1, B+N+1) - B-1;
        for(reg int i = 1; i <= N; i ++) A[i] = std::lower_bound(B+1, B+Len+1, A[i])-B;
        for(reg int i = 1; i <= Q_; i ++) que[i].l = read(), que[i].r = read(), que[i].id = i;
        int size = sqrt(N);
        for(reg int i = 1; i <= N; i ++){
                llim[++ bkcnt] = i, rlim[bkcnt] = std::min(N, i+size-1);
                for(; i <= rlim[bkcnt]; i ++) bkid[i] = bkcnt; i --;
        }
        std::sort(que+1, que+Q_+1, cmp);
        int lt = 1, rt = 0, last = 0;
        for(reg int i = 1; i <= Q_; i ++){
                if(bkid[que[i].l] == bkid[que[i].r]){ 
                        ll Tmp_1 = 0;
                        for(reg int j = que[i].l; j <= que[i].r; j ++) Tmp_1 = std::max(Tmp_1, 1ll*B[A[j]]*(++ Tong2[A[j]]));
                        for(reg int j = que[i].l; j <= que[i].r; j ++) Tong2[A[j]] --;
                        res[que[i].id] = Tmp_1; continue ;
                }
                if(last != bkid[que[i].l]){
                        while(lt < rlim[bkid[que[i].l]]+1) Del(lt ++);
                        while(rt > lt-1) Del(rt --);
                        Ans = 0; last = bkid[que[i].l];
                }
                while(rt < que[i].r) Add(++ rt); ll Tmp = Ans;
                while(lt > que[i].l) Add(-- lt); res[que[i].id] = Ans;
                while(lt < rlim[bkid[que[i].l]]+1) Del(lt ++); Ans = Tmp;
        }
        for(reg int i = 1; i <= Q_; i ++){ write(res[i]); putchar('
'); }
        return 0;
}