LOJ #2461. 「2018 集训队互测 Day 1」完美的队列（分块维护单调性）

题目
详见林旭恒《完美的队列》命题报告

转化为求每次区间加入后所有被加入的点被彻底弹出的最早时间。
然后发现区间相同插入时间不同时有单调性。
分块，把每个区间分为 $\sqrt n$ 个大块和 $\sqrt n$ 个单点。（一个询问变成 $\sqrt n$ 个询问）
对于每个大块和每个单点都分别用two pointers处理出每个询问的答案。
然后按颜色对时间区间取并，
按输入时间插入可以省掉排序，
时间 $O(n\sqrt n)$
MLE 85pts Code：

#include <bits/stdc++.h>
#define maxn 100005
#define S 100
#define mS (maxn / S) + 10
#define LL long long
using namespace std;

int n, m, a[maxn], bl[maxn], ed[maxn], L[maxn], R[maxn], X[maxn];
vector<tuple<int, int, int> > op[maxn];
vector<int> opa[mS];
int ar[S], addtag, Mx;
vector<pair<int, int> > G[maxn];
LL ans[maxn];

int main() {
    scanf("%d%d", &n, &m);
    memset(bl, -1, sizeof bl), bl[0] = 0;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), bl[i] = i / S;
    for (int i = 1; i <= m; i++) {
        int l, r, x;
        scanf("%d%d%d", &l, &r, &x);
        L[i] = l, R[i] = r, X[i] = x;
        for (int j = bl[l] + 1; j < bl[r]; j++) opa[j].push_back(i);
        if (bl[l] != bl[r]) {
            for (int j = l; j <= r && bl[j] == bl[l]; j++)
                op[j].push_back(make_tuple(opa[bl[j]].size(), x, i));
            for (int j = r; j >= l && bl[j] == bl[r]; j--)
                op[j].push_back(make_tuple(opa[bl[j]].size(), x, i));
        } else {
            for (int j = l; j <= r; j++) op[j].push_back(make_tuple(opa[bl[j]].size(), x, i));
        }
    }

    for (int i = 0; i * S <= n; i++) {
        Mx = addtag = 0;
        for (int j = 0; j < S; j++) Mx = max(Mx, ar[j] = (i * S + j <= n ? a[i * S + j] : 0));
        for (int x = 1, y = 2; x <= m;) {
            if (L[x] < i * S && (i + 1) * S - 1 < R[x]) {
                for (; Mx + addtag > 0 && y <= m; y++) {
                    if (L[y] < i * S && (i + 1) * S - 1 < R[y])
                        addtag--;
                    else if ((i * S <= L[y] && L[y] < (i + 1) * S) || (i * S <= R[y] && R[y] < (i + 1) * S)) {
                        Mx = -0x3f3f3f3f;
                        for (int j = 0; j < S; j++) {
                            if (L[y] - i * S <= j && j <= R[y] - i * S)
                                ar[j]--;
                            Mx = max(Mx, ar[j]);
                        }
                    }
                }
                ed[x] = max(ed[x], Mx + addtag > 0 ? m + 1 : y - 1);
            }
            ++x;
            if (x <= m) {
                if (L[x] < i * S && (i + 1) * S - 1 < R[x])
                    addtag++;
                else if ((i * S <= L[x] && L[x] < (i + 1) * S) || (i * S <= R[x] && R[x] < (i + 1) * S)) {
                    Mx = -0x3f3f3f3f;
                    for (int j = 0; j < S; j++) {
                        if (L[x] - i * S <= j && j <= R[x] - i * S)
                            ar[j]++;
                        Mx = max(Mx, ar[j]);
                    }
                }
            }
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int x = 0, y = 0, sz = op[i].size(); x < sz; x++) {
            for (; y < sz && y - x + get<0>(op[i][y]) - get<0>(op[i][x]) < a[i]; y++)
                ;
            if (y == sz && y - 1 - x + opa[bl[i]].size() - get<0>(op[i][x]) >= a[i])
                ed[get<2>(op[i][x])] =
                    max(ed[get<2>(op[i][x])], opa[bl[i]][a[i] + get<0>(op[i][x]) + x + 1 - y - 1]);
            else if (y == sz)
                ed[get<2>(op[i][x])] = max(ed[get<2>(op[i][x])], m + 1);
            else if (y - x + get<0>(op[i][y]) - get<0>(op[i][x]) == a[i])
                ed[get<2>(op[i][x])] = max(ed[get<2>(op[i][x])], get<2>(op[i][y]));
            else if (get<0>(op[i][y]) - (y - x + get<0>(op[i][y]) - get<0>(op[i][x]) - a[i]) >=
                     opa[bl[i]].size())
                ed[get<2>(op[i][x])] = max(ed[get<2>(op[i][x])], m + 1);
            else
                ed[get<2>(op[i][x])] =
                    max(ed[get<2>(op[i][x])],
                        opa[bl[i]][get<0>(op[i][y]) - (y - x + get<0>(op[i][y]) - get<0>(op[i][x]) - a[i])]);
        }
    }

    for (int i = 1; i <= m; i++) G[X[i]].push_back(make_pair(i, ed[i]));
    for (int i = 1; i <= 100000; i++) {
        int Max = 0;
        for (int j = 0, sz = G[i].size(); j < sz; j++) {
            if (G[i][j].first >= Max)
                ans[G[i][j].first]++, ans[G[i][j].second]--;
            else if (Max <= G[i][j].second)
                ans[Max]++, ans[G[i][j].second]--;
            Max = max(Max, G[i][j].second);
        }
    }
    for (int i = 1; i <= m; i++) ans[i] += ans[i - 1], printf("%d\n", ans[i]);
}

LOJ #2461. 「2018 集训队互测 Day 1」完美的队列（分块维护单调性）

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