LOJ #193 线段树历史和 Solution

Description

给定序列 $a=(a_1,a_2,\cdots ,a_n)$，另有序列 $h$，初始时 $h=a$.
有 $m$ 个操作分两种：

• $\mathrm{add}(l,r,k)$：对每个 $i\in [l,r]$ 执行 $a_i\leftarrow a_i+k$.
• $\mathrm{query}(l,r)$：求 ${\sum }_{i=l}^r{h}_i$.

每次 操作 后，对每个 $i\in [1,n]$ 执行 $h_i\leftarrow h_i+a_i$.

Limitations

$1\le n,m\le 10^5$
$1\le l\le r\le n$
$|a_i|,|k|\le 1000$
$1\text{s},512\text{MB}$

Solution

显然使用线段树，每个节点维护信息 $D=(\text{sum},\text{hsum},\text{len})$.
当然还需要标记 $T=(\text{tag},\text{htag},\text{upd})$：

• $\text{tag}$：$a_i$ 的加标记.
• $\text{htag}$：$h_i$ 的加标记.
• $\text{upd}$：$h_i$ 的更新次数.

$D+D$ 显然，$D+T$ 和 $T+T$ 可以用矩阵推，当然也可以直接靠理解：

• $\text{hsum}\leftarrow \text{hsum}+\text{sum}\times {\text{upd}}^{\prime }+{\text{htag}}^{\prime }\times \text{len}$（总共加了 $\text{upd}$ 次原序列，再加上自己的标记）
• $\text{sum}\leftarrow \text{sum}+{\text{tag}}^{\prime }\times \text{len}$（加上自己标记）
• $\text{htag}\leftarrow \text{htag}+\text{tag}\times {\text{upd}}^{\prime }+{\text{htag}}^{\prime }$（和 $\text{hsum}$ 同理）
• $\text{tag}\leftarrow \text{tag}+{\text{tag}}^{\prime }$
• $\text{upd}\leftarrow \text{upd}+{\text{upd}}^{\prime }$

然后套模板即可，时间复杂度 $O(m\log n)$.

Code

$2.8\text{KB},0.47\text{s},10.5\text{MB}\text{(in total, C++20 with O2)}$

#include <bits/stdc++.h>
using namespace std;using i64 = long long;
using ui64 = unsigned long long;
using i128 = __int128;
using ui128 = unsigned __int128;
using f4 = float;
using f8 = double;
using f16 = long double;template<class T>
bool chmax(T &a, const T &b){if(a < b){ a = b; return true; }return false;
}template<class T>
bool chmin(T &a, const T &b){if(a > b){ a = b; return true; }return false;
}namespace seg_tree {struct Node {int l, r, len;i64 sum, hissum, tag, histag, upd;};inline int ls(int u) { return 2 * u + 1; }inline int rs(int u) { return 2 * u + 2; }struct SegTree {vector<Node> tr;inline SegTree() {}inline SegTree(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) {tr[u].sum = tr[ls(mid)].sum + tr[rs(mid)].sum;tr[u].hissum = tr[ls(mid)].hissum + tr[rs(mid)].hissum;}inline void apply(int u, i64 tag, i64 histag, i64 upd) {tr[u].hissum += tr[u].sum * upd + histag * tr[u].len;tr[u].sum += tag * tr[u].len;tr[u].histag += tr[u].tag * upd + histag;tr[u].tag += tag;tr[u].upd += upd;}inline void pushdown(int u, int mid) {apply(ls(mid), tr[u].tag, tr[u].histag, tr[u].upd);apply(rs(mid), tr[u].tag, tr[u].histag, tr[u].upd);tr[u].tag = tr[u].histag = tr[u].upd = 0;}inline void build(int u, int l, int r, const vector<i64>& a) {tr[u].l = l, tr[u].r = r, tr[u].len = r - l + 1;if (l == r) return (void)(tr[u].sum = tr[u].hissum = a[l]);const int mid = (l + r) >> 1;build(ls(mid), l, mid, a);build(rs(mid), mid + 1, r, a);pushup(u, mid);}inline void add(int u, int l, int r, i64 k) {if (l <= tr[u].l && tr[u].r <= r) return apply(u, k, 0, 0);const int mid = (tr[u].l + tr[u].r) >> 1;pushdown(u, mid);if (l <= mid) add(ls(mid), l, r, k);if (r > mid) add(rs(mid), l, r, k);pushup(u, mid);}inline i64 query(int u, int l, int r) {if (l <= tr[u].l && tr[u].r <= r) return tr[u].hissum;const int mid = (tr[u].l + tr[u].r) >> 1;i64 res = 0;pushdown(u, mid);if (l <= mid) res += query(ls(mid), l, r);if (r > mid) res += query(rs(mid), l, r);return res;}inline void range_add(int l, int r, i64 k) { add(0, l, r, k); }inline i64 range_hsum(int l, int r) { return query(0, l, r); }};
}
using seg_tree::SegTree;signed main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int n, m;scanf("%d %d", &n, &m);vector<i64> a(n);for (int i = 0; i < n; i++) scanf("%lld", &a[i]);SegTree sgt(a);for (int i = 0, op, l, r, v; i < m; i++) {scanf("%d %d %d", &op, &l, &r), l--, r--;if (op == 1) sgt.range_add(l, r, (scanf("%d", &v), v));else printf("%lld\n", sgt.range_hsum(l, r));sgt.apply(0, 0, 0, 1);}return 0;
}