P7492 [传智杯 #3 决赛] 序列 Solution-优快云博客

Description

给定长为 $n$ 的序列 $a=(a_1,a_2,\cdots,a_n)$ ，有 $m$ 个操作，分以下两种：

$\operatorname{query}(l,r)$ ：求 $\max(0,\max\limits_{[u,v] \in [l,r]}\sum\limits_{i=u}^v a_i)$ 。
$\operatorname{modify}(l,r,k)$ ：对于每个 $i\in [l,r]$ 执行 $a_i \leftarrow a_i \operatorname{or} k$ ，其中 $\operatorname{or}$ 为按位或。

Limitations

$\le n, m \le 10^5$
$|a_i|,|k| \le 2^{30}$
$\le l \le r \le n$
$1\text{s},128\text{MB}$

Solution

$\operatorname{modify}$ 操作无法打标记，考虑暴力修改，但直接暴力改会 $\text{TLE}$ .
考虑对每个区间维护 $\operatorname{and}$ 和，当发现区间 $\operatorname{and}$ 和 $\operatorname{or}$ 上 $k$ 不变，那么继续修改显然不会改变 $a_i$ ，可以不向该区间递归.
剩下的就是 P4513 了.
时间复杂度 $O(n\log n\log V)$ .

Code

$3.17\text{KB},5.87\text{s},10.4\text{MB} \; (\texttt{in total, C++20 with O2})$
重构过了.

// Problem: P7492 [传智杯 #3 决赛] 序列
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P7492
// Memory Limit: 128 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#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;
	    i64 lmax, rmax, tsum, sum, andd;
	};
	
	inline void merge(Node& res, const Node& le, const Node& ri) {
		res.lmax = max(le.lmax, le.sum + ri.lmax);
		res.rmax = max(le.rmax + ri.sum, ri.rmax);
		res.tsum = max({le.tsum, ri.tsum, le.rmax + ri.lmax});
		res.sum = le.sum + ri.sum;
		res.andd = le.andd & ri.andd;
	}
	
	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) { 
			merge(tr[u], tr[ls(mid)], tr[rs(mid)]); 
		}
		
		inline void build(int u, int l, int r, const vector<i64>& a) {
		    tr[u].l = l;
		    tr[u].r = r;
		    if (l == r) {
		        tr[u].sum = tr[u].andd = a[l];
		        tr[u].lmax = tr[u].rmax = tr[u].tsum = max(a[l], 0LL);
		        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 update(int u, int l, int r, i64 val) {
		    if (tr[u].l == tr[u].r) {
		        tr[u].sum |= val;
		        tr[u].andd = tr[u].sum;
		        tr[u].lmax = tr[u].rmax = tr[u].tsum = max(tr[u].sum, 0LL);
		        return;
		    }
		    const int mid = (tr[u].l + tr[u].r) >> 1;
		    if (l <= mid && (tr[ls(mid)].andd | val) != tr[ls(mid)].andd) update(ls(mid), l, r, val);
		    if (r > mid && (tr[rs(mid)].andd | val) != tr[rs(mid)].andd) update(rs(mid), l, r, val);
		    pushup(u, mid);
		}
		
		inline Node query(int u, int l, int r) {
		    if (l <= tr[u].l && tr[u].r <= r) {
		        return tr[u];
		    }
		    const int mid = (tr[u].l + tr[u].r) >> 1;
		    if (r <= mid) return query(ls(mid), l, r);
		    if (l > mid) return query(rs(mid), l, r);
		    Node le = query(ls(mid), l, r), ri = query(rs(mid), l, r), res;
		    merge(res, le, ri);
		    return res;
		}
		
		inline void range_or(int l, int r, i64 x) { update(0, l, r, x); }
		inline i64 range_gss(int l, int r) { return query(0, l, r).tsum; }
	};
}
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) printf("%lld\n", sgt.range_gss(l, r));
	    else {
	        scanf("%d", &v);
	        sgt.range_or(l, r, v);
	    }
	}
	return 0;
}