P8512 [Ynoi Easy Round 2021] TEST_152 Solution

Description

有一序列 $c=(c_1,c_2,\cdots ,c_m)$ 和 $n$ 个三元组 $(l_i,r_i,v_i)$.
回答 $q$ 次形如 $(L,R)$ 的询问，具体如下：

• 将 $c$ 置为全 $0$.
• 对每个 $i\in [L,R]$，将 $c_{l_i}\sim c_{r_i}$ 赋值为 $v_i$.
• 最后输出 ${\sum }_{i=1}^m{c}_i$.

询问之间互相独立.

Limitations

$1\le n,m,q\le 5\times 10^5$
$1\le l_i\le r_i\le m$
$1\le L\le R\le n$

Solution

典题一道.
答案不好在线求，但可由前缀相减得到，考虑挂在 $r$ 上做扫描线，并开 BIT 维护时间轴.
我们都知道，区间 $[l,r]$ 覆盖 $v$ 后的贡献为 $v\times (r-l+1)$.
然而覆盖后需要抹除之前的贡献，而这可以借助 ODT 完成.
具体地，在 assign 时将 $[l,r]$ 中原有的颜色段的贡献在 BIT 上一段段减掉.
然后就做完了，代码很好写.
时间复杂度 $O(n\log n)$，具体请自证.

Code

$2.65\text{KB},10.01\text{s},64.81\text{MB}\text{(in total, C++20 with O2)}$

#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;
}struct Node {int l, r;mutable int v;int t;inline Node(int l, int r = 0, int v = 0, int t = 0) : l(l), r(r), v(v), t(t) {}inline bool operator<(const Node& rhs) const { return l < rhs.l; }
};struct ODT {set<Node> s;using Iter = set<Node>::iterator;inline ODT() {}inline ODT(int n) {s.emplace(0, n - 1, 0, 0);}inline Iter split(int pos) {auto it = s.lower_bound(Node(pos, 0, 0, 0));if (it != s.end() && it->l == pos) return it;--it;if (it->r < pos) return s.end();auto [l, r, val, tim] = *it;s.erase(it), s.emplace(l, pos - 1, val, tim);return s.emplace(pos, r, val, tim).first;}inline void assign(int l, int r, int v, int t, vector<Node>& opt) {opt.clear();auto itr = split(r + 1), itl = split(l);for (auto it = itl; it != itr; it++) opt.push_back(*it);s.erase(itl, itr);s.emplace(l, r, v, t);}
};int lowbit(int x){return x & -x;
}template<class T>
struct fenwick{int n;vector<T> c;fenwick() {}fenwick(int _n): n(_n){c.resize(n + 1);}fenwick(const vector<T> &a): n(a.size()){c.resize(n + 1);for(int i = 1; i <= n; i++){c[i] = c[i] + a[i - 1];int j = i + lowbit(i);if(j <= n) c[j] = c[j] + c[i];}}void add(int x, const T& v){for(int i = x + 1; i <= n; i += lowbit(i)) c[i] = c[i] + v;}T ask(int x){T ans{};for(int i = x + 1; i; i -= lowbit(i)) ans = ans + c[i];return ans;}T ask(int l, int r){return ask(r) - ask(l - 1);}
};using pii = pair<int, int>;signed main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int n, m, q;scanf("%d %d %d", &n, &m, &q);vector<int> L(n), R(n), X(n);for (int i = 0; i < n; i++) {scanf("%d %d %d", &L[i], &R[i], &X[i]);L[i]--, R[i]--;}vector<vector<pii>> adj(n);for (int i = 0, l, r; i < q; i++) {scanf("%d %d", &l, &r), l--, r--;adj[r].emplace_back(l, i);}ODT odt(m);fenwick<i64> fwk(n);vector<Node> tmp;vector<i64> ans(q);for (int i = 0; i < n; i++) {odt.assign(L[i], R[i], X[i], i, tmp);for (auto [l, r, v, t] : tmp) fwk.add(t, -1LL * (r - l + 1) * v);fwk.add(i, 1LL * (R[i] - L[i] + 1) * X[i]);for (auto [j, id] : adj[i]) ans[id] = fwk.ask(j, i);}for (int i = 0; i < q; i++) printf("%lld\n", ans[i]);return 0;
}