Japanese Student Championship 2021 F Max Matrix

洛谷传送门

AtCoder 传送门

我们考虑动态维护答案。如果已经知道上一步的答案，考虑计算这一次操作之后答案的变化。

考虑现在有一个数列 \(a_{1 \sim n}\)，我们想知道 \(\sum\limits_{i = 1}^n \max(x, a_i)\)。

计算出 \(c = \sum\limits_{i = 1}^n [a_i < x], s = \sum\limits_{i = 1}^n [a_i \ge x] a_i\)，答案显然就是 \(s + c \times x\)。

这题和上面的题本质相同。以第一种操作为例，每次答案减去旧的 \(a_x\) 对 \(b\) 造成的贡献，再加上 \(y\) 对 \(b\) 造成的贡献。第二种操作是对称的。

于是我们只需要对 \(y_i\) 离散化后，行和列分别维护两个树状数组表示 \(\sum\limits_{i = 1}^n [a_i \le x]\) 和 \(\sum\limits_{i = 1}^n [a_i \le x] a_i\)，一共四个树状数组即可。

时间复杂度 \(O(q \log (q + n + m))\)。

code

// Problem: F - Max Matrix
// Contest: AtCoder - Japanese Student Championship 2021
// URL: https://atcoder.jp/contests/jsc2021/tasks/jsc2021_f
// Memory Limit: 1024 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#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 = 200100;

ll n, m, q, a[maxn], b[maxn], lsh[maxn], tot;

struct node {
	ll op, x, y;
} c[maxn];

struct BIT {
	ll c[maxn];
	
	inline void update(int x, ll d) {
		for (int i = x; i <= tot; i += (i & (-i))) {
			c[i] += d;
		}
	}
	
	inline ll query(int x) {
		ll res = 0;
		for (int i = x; i; i -= (i & (-i))) {
			res += c[i];
		}
		return res;
	}
	
	inline ll query(int l, int r) {
		return query(r) - query(l - 1);
	}
} t1, t2, t3, t4;

inline ll calc1(ll x) {
	return t4.query(x, tot) + lsh[x] * t3.query(x - 1);
}

inline ll calc2(ll x) {
	return t2.query(x, tot) + lsh[x] * t1.query(x - 1);
}

void solve() {
	scanf("%lld%lld%lld", &n, &m, &q);
	lsh[++tot] = 0;
	for (int i = 1; i <= q; ++i) {
		scanf("%lld%lld%lld", &c[i].op, &c[i].x, &c[i].y);
		lsh[++tot] = c[i].y;
	}
	sort(lsh + 1, lsh + tot + 1);
	tot = unique(lsh + 1, lsh + tot + 1) - lsh - 1;
	for (int i = 1; i <= n; ++i) {
		a[i] = 1;
	}
	for (int i = 1; i <= m; ++i) {
		b[i] = 1;
	}
	t1.update(1, n);
	t3.update(1, m);
	ll ans = 0;
	for (int i = 1; i <= q; ++i) {
		ll op = c[i].op, x = c[i].x, y = c[i].y;
		y = lower_bound(lsh + 1, lsh + tot + 1, y) - lsh;
		if (op == 1) {
			t1.update(a[x], -1);
			t2.update(a[x], -lsh[a[x]]);
			ans -= calc1(a[x]);
			a[x] = y;
			t1.update(y, 1);
			t2.update(y, lsh[y]);
			ans += calc1(y);
		} else {
			t3.update(b[x], -1);
			t4.update(b[x], -lsh[b[x]]);
			ans -= calc2(b[x]);
			b[x] = y;
			t3.update(y, 1);
			t4.update(y, lsh[y]);
			ans += calc2(y);
		}
		printf("%lld\n", ans);
	}
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}