Description
给定序列 a = ( a 1 , a 2 , ⋯ , a n ) a=(a_1,a_2,\cdots,a_n) a=(a1,a2,⋯,an),有 m m m 个操作分两种:
- chmax ( l , r , v ) \operatorname{chmax}(l,r,v) chmax(l,r,v):对每个 i ∈ [ l , r ] i \in [l,r] i∈[l,r] 执行 a i ← max ( a i , v ) a_i \gets \max(a_i,v) ai←max(ai,v).
- query ( l , r ) \operatorname{query}(l,r) query(l,r):求 max ( 0 , max [ u , v ] ∈ [ l , r ] ∑ i = u v a i ) \max(0,\max\limits_{[u,v]\in [l,r]}\sum\limits_{i=u}^v a_i) max(0,[u,v]∈[l,r]maxi=u∑vai).
Limitations
1
≤
n
≤
10
5
1 \le n \le 10^5
1≤n≤105
1
≤
m
≤
2
×
10
5
1 \le m \le 2\times 10^5
1≤m≤2×105
1
≤
l
≤
r
≤
n
1 \le l \le r \le n
1≤l≤r≤n
∣
a
i
∣
,
∣
v
∣
≤
10
9
|a_i|,|v| \le 10^9
∣ai∣,∣v∣≤109
3
s
,
512
MB
3\text{s},512\text{MB}
3s,512MB
Solution
看到
chmax
\operatorname{chmax}
chmax 先来一个吉司机.
然后下传标记相当于区间加上
(
v
−
min
)
(v-\min)
(v−min),这样就把问题变成区间加正数.
然后可以上 KTT 维护,具体见 P5693(这里把
k
k
k 视作最小值的出现次数),要注意 pushup 时将两者间较大值的
k
k
k 清零.
两套势能系统互相独立,故时间复杂度为
O
(
(
n
+
m
)
log
3
n
)
O((n+m)\log^3n)
O((n+m)log3n).
Code
只给一部分.
4.92
KB
,
4.71
s
,
21.16
MB
(in
total,
C++20
with
O2)
4.92\text{KB},4.71\text{s},21.16\text{MB}\;\texttt{(in total, C++20 with O2)}
4.92KB,4.71s,21.16MB(in total, C++20 with O2)
constexpr i64 inf = 3e18;
struct line {
int k; i64 b;
inline line() : line(0, 0) {}
inline line(int _k, i64 _b): k(_k), b(_b) {}
inline void add(i64 v) { b += k * v; }
inline void reset() { k = 0; }
};
inline line operator+(const line& lhs, const line& rhs) {
return line(lhs.k + rhs.k, lhs.b + rhs.b);
}
inline pair<line, i64> _max(const line& a, const line& b) {
if (a.k < b.k || (a.k == b.k && a.b < b.b)) return _max(b, a);
if (a.b >= b.b) return make_pair(a, inf);
return make_pair(b, (b.b - a.b) / (a.k - b.k));
}
struct info {
line pre, suf, ans, sum;
i64 x;
inline info() {}
inline info(line pre, line suf, line ans, line sum, i64 x)
: pre(pre), suf(suf), ans(ans), sum(sum), x(x) {}
inline info reset() {
info res = *this;
res.pre.reset();
res.suf.reset();
res.ans.reset();
res.sum.reset();
return res;
}
};
inline info operator+(const info& a, const info& b) {
i64 x0 = min(a.x, b.x);
line sum = a.sum + b.sum;
auto [pre, x1] = _max(a.pre, a.sum + b.pre);
auto [suf, x2] = _max(b.suf, a.suf + b.sum);
auto [tmp, x3] = _max(a.ans, b.ans);
auto [ans, x4] = _max(tmp, a.suf + b.pre);
return info(pre, suf, ans, sum, min({x0, x1, x2, x3, x4}));
}
inline void operator+=(info& a, i64 v) {
a.x -= v;
a.pre.add(v), a.suf.add(v);
a.sum.add(v), a.ans.add(v);
}
struct node {
int l, r;
info dat;
i64 tag, min, sec;
};
inline int ls(int u) { return 2 * u + 1; }
inline int rs(int u) { return 2 * u + 2; }
struct ktt {
vector<node> tr;
inline ktt() {}
inline ktt(const vector<i64>& a) {
const int n = a.size();
tr.resize(n << 1);
build(0, 0, n - 1, a);
}
inline void pushup(int u, int mid) {
if (tr[ls(mid)].min == tr[rs(mid)].min) {
tr[u].min = tr[ls(mid)].min;
tr[u].sec = min(tr[ls(mid)].sec, tr[rs(mid)].sec);
tr[u].dat = tr[ls(mid)].dat + tr[rs(mid)].dat;
}
else if (tr[ls(mid)].min < tr[rs(mid)].min) {
tr[u].min = tr[ls(mid)].min;
tr[u].sec = min(tr[ls(mid)].sec, tr[rs(mid)].min);
tr[u].dat = tr[ls(mid)].dat + tr[rs(mid)].dat.reset();
}
else {
tr[u].min = tr[rs(mid)].min;
tr[u].sec = min(tr[ls(mid)].min, tr[rs(mid)].sec);
tr[u].dat = tr[ls(mid)].dat.reset() + tr[rs(mid)].dat;
}
}
inline void apply(int u, i64 w) {
if (w <= tr[u].min) return;
i64 v = w - tr[u].min;
tr[u].min = w;
tr[u].tag = max(tr[u].tag, w);
tr[u].dat += v;
}
inline void pushdown(int u, int mid) {
if (tr[u].tag != -inf) {
apply(ls(mid), tr[u].tag);
apply(rs(mid), tr[u].tag);
tr[u].tag = -inf;
}
}
inline void build(int u, int l, int r, const vector<i64>& a) {
tr[u].l = l, tr[u].r = r, tr[u].tag = -inf;
if (l == r) {
line f(1, a[l]);
tr[u].dat = info(f, f, f, f, inf);
tr[u].min = a[l], tr[u].sec = inf;
return;
}
const int mid = (l + r) >> 1;
build(ls(mid), l, mid, a);
build(rs(mid), mid + 1, r, a);
pushup(u, mid);
}
inline void defeat(int u, i64 v) {
tr[u].tag = max(tr[u].tag, v);
if (v - tr[u].min > tr[u].dat.x) {
const int mid = (tr[u].l + tr[u].r) >> 1;
defeat(ls(mid), v);
defeat(rs(mid), v);
pushup(u, mid);
}
else apply(u, v);
}
inline void update(int u, int l, int r, i64 v) {
if (tr[u].min >= v) return;
if (l <= tr[u].l && tr[u].r <= r && v < tr[u].sec) return defeat(u, v);
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (l <= mid) update(ls(mid), l, r, v);
if (r > mid) update(rs(mid), l, r, v);
pushup(u, mid);
}
inline info query(int u, int l, int r) {
if (l <= tr[u].l && tr[u].r <= r) return tr[u].dat;
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (r <= mid) return query(ls(mid), l, r);
if (l > mid) return query(rs(mid), l, r);
return query(ls(mid), l, r) + query(rs(mid), l, r);
}
inline void range_chmax(int l, int r, i64 v) { update(0, l, r, v); }
inline i64 range_gss(int l, int r) { return max(query(0, l, r).ans.b, 0LL); }
};

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