UOJ169. 【UR #11】元旦老人与数列

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传送门
考虑用 \(segment~tree~beats\) 那一套理论，维护区间最小值 \(mn\) 和严格次小值 \(se\)
那么可以直接 \(mlog^2n\) 维护前三个操作
考虑维护历史最小值，先维护历史最小标记
写了写发现 \(max\) 那个修改不好操作
对于 \(max\) 操作来说，只会在 \(mn< v <se\) 的时候打上标记
这就相当于区间内等于 \(mn\) 的权值都要变成 \(v\)
那么 \(max\) 操作就可以变成对区间最小值的加法操作
而 \(v<se\)，这样就可以非常方便维护历史最小值了
具体来说，维护下面几个标记

1. 区间最小值的加法标记
2. 区间其它值的加法标记
3. 区间最小值的历史最小的加法标记
4. 区间其它值的历史最小的加法标记
   下放的时候判断一下是否是区间最小值就好了

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

namespace IO {
    const int maxn(1 << 21 | 1);

    char obuf[maxn], ibuf[maxn], *iS, *iT, c, *oS = obuf, *oT = obuf + maxn - 1, st[60];
    int f, tp;
    
    inline char Getc() {
        return iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), (iS == iT ? EOF : *iS++)) : *iS++;
    }

    template <class Int> inline void In(Int &x) {
        for (c = Getc(), f = 1; c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;
        for (x = 0; c >= '0' && c <= '9'; c = Getc()) x = (x << 1) + (x << 3) + (c ^ 48);
        x *= f;
    }

    inline void Flush() {
        fwrite(obuf, 1, oS - obuf, stdout);
        oS = obuf;
    }

    inline void Putc(char c) {
        *oS++ = c;
        if (oS == oT) Flush();
    }

    template <class Int> inline void Out(Int x) {
        if (x < 0) Putc('-'), x = -x;
        if (!x) Putc('0');
        while (x) st[++tp] = x % 10 + '0', x /= 10;
        while (tp) Putc(st[tp--]);
    }
}

using IO :: In;
using IO :: Out;
using IO :: Putc;
using IO :: Flush;

const int maxn(2e6 + 5);
const int inf(2e9);

struct Min {
    int mn1, mn2;

    inline Min operator +(Min b) const {
        Min c;
        c.mn1 = min(mn1, b.mn1), c.mn2 = min(mn2, b.mn2);
        if (b.mn1 ^ c.mn1) c.mn2 = min(c.mn2, b.mn1);
        if (mn1 ^ c.mn1) c.mn2 = min(c.mn2, mn1);
        return c;
    }
};

Min mn[maxn];
int n, m, hmn[maxn], addmn1[maxn], addmn2[maxn], addhmn1[maxn], addhmn2[maxn];

inline void Update(int x) {
    mn[x] = mn[x << 1] + mn[x << 1 | 1], hmn[x] = min(hmn[x << 1], hmn[x << 1 | 1]);
}

void Build(int x, int l, int r) {
    int mid;
    if (l == r) {
        In(mn[x].mn1), mn[x].mn2 = inf, hmn[x] = mn[x].mn1;
        return;
    }
    mid = (l + r) >> 1;
    Build(x << 1, l, mid), Build(x << 1 | 1, mid + 1, r);
    Update(x);
}

inline void Puttag(int x, int vmn1, int vmn2, int vhmn1, int vhmn2) {
    hmn[x] = min(hmn[x], mn[x].mn1 + vhmn1);
    addhmn1[x] = min(addhmn1[x], addmn1[x] + vhmn1);
    addhmn2[x] = min(addhmn2[x], addmn2[x] + vhmn2);
    addmn1[x] += vmn1, addmn2[x] += vmn2, mn[x].mn1 += vmn1;
    if (mn[x].mn2 ^ inf) mn[x].mn2 += vmn2;
}

inline void Pushdown(int x) {
    if (!addmn1[x] && !addmn2[x] && !addhmn1[x] && !addhmn2[x]) return;
    int ls, rs, now;
    ls = x << 1, rs = x << 1 | 1, now = min(mn[ls].mn1, mn[rs].mn1);
    if (now == mn[ls].mn1) Puttag(ls, addmn1[x], addmn2[x], addhmn1[x], addhmn2[x]);
    else Puttag(ls, addmn2[x], addmn2[x], addhmn2[x], addhmn2[x]);
    if (now == mn[rs].mn1) Puttag(rs, addmn1[x], addmn2[x], addhmn1[x], addhmn2[x]);
    else Puttag(rs, addmn2[x], addmn2[x], addhmn2[x], addhmn2[x]);
    addmn1[x] = addmn2[x] = addhmn1[x] = addhmn2[x] = 0;
}

void Modify_add(int x, int l, int r, int ql, int qr, int v) {
    int mid;
    if (ql <= l && qr >= r) {
        Puttag(x, v, v, v, v);
        return;
    }
    mid = (l + r) >> 1, Pushdown(x);
    if (ql <= mid) Modify_add(x << 1, l, mid, ql, qr, v);
    if (qr > mid) Modify_add(x << 1 | 1, mid + 1, r, ql, qr, v);
    Update(x);
}

void Modify_max(int x, int l, int r, int ql, int qr, int v) {
    int mid;
    if (mn[x].mn1 >= v) return;
    if (ql <= l && qr >= r && mn[x].mn2 > v) {
        Puttag(x, v - mn[x].mn1, 0, v - mn[x].mn1, 0);
        return;
    }
    mid = (l + r) >> 1, Pushdown(x);
    if (ql <= mid) Modify_max(x << 1, l, mid, ql, qr, v);
    if (qr > mid) Modify_max(x << 1 | 1, mid + 1, r, ql, qr, v);
    Update(x);
}

int Query_min(int x, int l, int r, int ql, int qr) {
    int mid, ret;
    if (ql <= l && qr >= r) return mn[x].mn1;
    mid = (l + r) >> 1, Pushdown(x), ret = inf;
    if (ql <= mid) ret = Query_min(x << 1, l, mid, ql, qr);
    if (qr > mid) ret = min(ret, Query_min(x << 1 | 1, mid + 1, r, ql, qr));
    Update(x);
    return ret;
}

int Query_hmin(int x, int l, int r, int ql, int qr) {
    int mid, ret;
    if (ql <= l && qr >= r) return hmn[x];
    mid = (l + r) >> 1, Pushdown(x), ret = inf;
    if (ql <= mid) ret = Query_hmin(x << 1, l, mid, ql, qr);
    if (qr > mid) ret = min(ret, Query_hmin(x << 1 | 1, mid + 1, r, ql, qr));
    Update(x);
    return ret;
}

int main() {
    int i, op, l, r, v;
    In(n), In(m), Build(1, 1, n);
    while (m) {
        --m, In(op), In(l), In(r);
        if (op == 1) In(v), Modify_add(1, 1, n, l, r, v);
        else if (op == 2) In(v), Modify_max(1, 1, n, l, r, v);
        else if (op == 3) Out(Query_min(1, 1, n, l, r)), Putc('\n');
        else Out(Query_hmin(1, 1, n, l, r)), Putc('\n');
    }
    return Flush(), 0;
}

UOJ169. 【UR #11】元旦老人与数列

标签：put   DDM   str   write   tor   ref   span   amp   while   

原文地址：https://www.cnblogs.com/cjoieryl/p/10251754.html