CodeForces 1924B Space Harbour

洛谷传送门

CF 传送门

不知道为什么好像大家在这题上花了挺久的。

发现对于一对相邻的港口 $(x_i, x_{i + 1})$，$x \in (x_i, x_{i + 1})$ 的花费是 $y_i (x_{i + 1} - x)$。拆开得 $y_i x_{i + 1} - y_i x$。

考虑用 set 维护所有港口，这样可以知道一个港口左边和右边的港口的坐标和价值。那么前一项 $y_i x_{i + 1}$ 可以线段树区间覆盖，后一项 $-y_i x$ 也可以线段树区间覆盖处理，相当于让一个代表 $[l, r]$ 区间的线段树结点的和变成 $- y_i \frac{(l + r)(r - l + 1)}{2}$。这个标记是可以下传的。

总时间复杂度为 $O((m + q) \log n)$。

code

// Problem: B. Space Harbour
// Contest: Codeforces - Codeforces Round 921 (Div. 1)
// URL: https://codeforces.com/contest/1924/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)
 
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
 
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
 
const int maxn = 300100;
 
ll n, m, q, a[maxn], b[maxn];
 
inline ll calc(ll l, ll r) {
	return (l + r) * (r - l + 1) / 2;
}
 
struct {
	ll a[maxn << 2], tag[maxn << 2];
	
	inline void pushup(int x) {
		a[x] = a[x << 1] + a[x << 1 | 1];
	}
	
	inline void init() {
		mems(tag, -1);
	}
	
	inline void pushdown(int x, int l, int r) {
		if (tag[x] == -1) {
			return;
		}
		int mid = (l + r) >> 1;
		a[x << 1] = tag[x] * (mid - l + 1);
		a[x << 1 | 1] = tag[x] * (r - mid);
		tag[x << 1] = tag[x << 1 | 1] = tag[x];
		tag[x] = -1;
	}
	
	void update(int rt, int l, int r, int ql, int qr, ll x) {
		if (ql > qr) {
			return;
		}
		if (ql <= l && r <= qr) {
			a[rt] = x * (r - l + 1);
			tag[rt] = x;
			return;
		}
		pushdown(rt, l, r);
		int mid = (l + r) >> 1;
		if (ql <= mid) {
			update(rt << 1, l, mid, ql, qr, x);
		}
		if (qr > mid) {
			update(rt << 1 | 1, mid + 1, r, ql, qr, x);
		}
		pushup(rt);
	}
	
	ll query(int rt, int l, int r, int ql, int qr) {
		if (ql <= l && r <= qr) {
			return a[rt];
		}
		pushdown(rt, l, r);
		int mid = (l + r) >> 1;
		ll res = 0;
		if (ql <= mid) {
			res += query(rt << 1, l, mid, ql, qr);
		}
		if (qr > mid) {
			res += query(rt << 1 | 1, mid + 1, r, ql, qr);
		}
		return res;
	}
} t1;
 
struct {
	ll a[maxn << 2], tag[maxn << 2];
	
	inline void pushup(int x) {
		a[x] = a[x << 1] + a[x << 1 | 1];
	}
	
	inline void init() {
		mems(tag, -1);
	}
	
	inline void pushdown(int x, int l, int r) {
		if (tag[x] == -1) {
			return;
		}
		int mid = (l + r) >> 1;
		a[x << 1] = tag[x] * calc(l, mid);
		a[x << 1 | 1] = tag[x] * calc(mid + 1, r);
		tag[x << 1] = tag[x << 1 | 1] = tag[x];
		tag[x] = -1;
	}
	
	void update(int rt, int l, int r, int ql, int qr, ll x) {
		if (ql > qr) {
			return;
		}
		if (ql <= l && r <= qr) {
			a[rt] = x * calc(l, r);
			tag[rt] = x;
			return;
		}
		pushdown(rt, l, r);
		int mid = (l + r) >> 1;
		if (ql <= mid) {
			update(rt << 1, l, mid, ql, qr, x);
		}
		if (qr > mid) {
			update(rt << 1 | 1, mid + 1, r, ql, qr, x);
		}
		pushup(rt);
	}
	
	ll query(int rt, int l, int r, int ql, int qr) {
		if (ql <= l && r <= qr) {
			return a[rt];
		}
		pushdown(rt, l, r);
		int mid = (l + r) >> 1;
		ll res = 0;
		if (ql <= mid) {
			res += query(rt << 1, l, mid, ql, qr);
		}
		if (qr > mid) {
			res += query(rt << 1 | 1, mid + 1, r, ql, qr);
		}
		return res;
	}
} t2;
 
set<pii> S;
 
inline void upd(ll x, ll y, ll z) {
	t1.update(1, 1, n, x + 1, y - 1, z * y);
	t2.update(1, 1, n, x + 1, y - 1, -z);
}
 
void solve() {
	t1.init();
	t2.init();
	scanf("%lld%lld%lld", &n, &m, &q);
	for (int i = 1; i <= m; ++i) {
		scanf("%lld", &a[i]);
	}
	for (int i = 1; i <= m; ++i) {
		scanf("%lld", &b[i]);
		S.emplace(a[i], b[i]);
	}
	for (auto it = S.begin(); next(it) != S.end(); ++it) {
		auto jt = next(it);
		upd(it->fst, jt->fst, it->scd);
	}
	while (q--) {
		ll op, x, y;
		scanf("%lld%lld%lld", &op, &x, &y);
		if (op == 1) {
			S.emplace(x, y);
			t1.update(1, 1, n, x, x, 0);
			t2.update(1, 1, n, x, x, 0);
			auto it = S.find(mkp(x, y));
			auto p = prev(it), q = next(it);
			upd(p->fst, it->fst, p->scd);
			upd(it->fst, q->fst, it->scd);
		} else {
			printf("%lld\n", t1.query(1, 1, n, x, y) + t2.query(1, 1, n, x, y));
		}
	}
}
 
int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}