Submission #1074158

#	Time	Username	Problem	Language	Result	Execution time	Memory
1074158		Gromp15	Catfish Farm (IOI22_fish)	C++17	12 / 100	1032 ms	2097152 KiB

fish

#include <bits/stdc++.h>
#include "fish.h"
#define ll long long
#define ar array
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
using namespace std;
 
const ll INF = 1e18;
template<typename T> bool ckmin(T &a, const T &b ) { return a > b ? a = b, 1 : 0; }
template<typename T> bool ckmax(T &a, const T &b ) { return a < b ? a = b, 1 : 0; }
 
struct seg {
	int N; vector<ll> tree;
	seg(int n) : N(1<<(__lg(n)+1)), tree(2*N, -1e18) {}
	void update(int pos, ll x) {
		for (int i = pos + N; i; i >>= 1) ckmax(tree[i], x);
	}
	ll query(int node, int nl, int nr, int ql, int qr) {
		if (ql > nr || qr < nl) return -1e18;
		if (ql <= nl && nr <= qr) return tree[node];
		int mid = (nl+nr)/2;
		return max(query(node*2, nl, mid, ql, qr), query(node*2+1, mid+1, nr, ql, qr));
	}
	ll query(int l, int r) {
		return query(1, 0, N-1, l, r);
	}
};

long long max_weights(int n, int m, std::vector<int> x, std::vector<int> y, std::vector<int> w) {
	vector<vector<ar<ll, 2>>> each(n);
	for (int i = 0; i < m; i++) {
		each[x[i]].push_back({y[i], w[i]});
		each[x[i]].push_back({y[i]-1, 0});
	}
	for (int i = 0; i < n; i++) each[i].push_back({n-1, 0}), each[i].push_back({-1, 0});
	for (int i = 0; i < n; i++) {
		sort(all(each[i]));
		vector<ar<ll, 2>> nw;
		ll s = 0;
		for (int j = 0; j < sz(each[i]); j++) {
			int r = j;
			while (r+1 < sz(each[i]) && each[i][r+1][0] == each[i][r][0]) r++;
			for (int k = j; k <= r; k++) s += each[i][k][1];
			nw.push_back({each[i][j][0], s});
			j = r;
		}
		assert(is_sorted(all(nw)));
		swap(each[i], nw);
	}
	vector<vector<ll>> dp;
	auto query = [&](int pos, int x) {
		return prev(upper_bound(all(each[pos]), ar<ll, 2>{x, LLONG_MAX})) - each[pos].begin();
	};
	vector<vector<ar<int, 3>>> val(n);
	for (int i = 0; i < n; i++) {
		val[i].resize(sz(each[i]));
		for (int k = 0; k < sz(each[i]); k++) {
			for (int j = -1; j <= 1; j++) {
				if (i + j >= 0 && i + j < n) {
					val[i][k][j+1] = query(i + j, each[i][k][0]);
				}
			}
		}
	}
	vector<vector<ll>> dp3, dp4, dp5;
	for (int i = 0; i < n; i++) {
		vector<vector<ll>> dp2;
		const int N = sz(each[i]);
		dp2.resize(i ? sz(each[i-1]) : 1, vector<ll>(N, -INF));
		if (!i) {
			for (int j = 0; j < N; j++) {
				dp2[0][j] = 0;
			}
			swap(dp, dp2);
		}
		else {
			dp3 = dp, dp4 = dp;
			for (int i = 1; i < sz(dp3); i++) for (int j = 0; j < sz(dp[i]); j++) ckmax(dp3[i][j], dp3[i-1][j]);
			for (int j = sz(dp4)-1; j >= 0; j--) for (int k = 0; k < sz(dp[j]); k++) dp4[j][k] = dp[j][k] + (i-2 >= 0 ? each[i-1][val[i-2][j][2]][1] : 0) - each[i-1][k][1];
			for (int j = sz(dp4)-2; j >= 0; j--) for (int k = 0; k < sz(dp4[j]); k++) ckmax(dp4[j][k], dp4[j+1][k]);
			vector<seg> st(sz(dp[0]), seg(sz(dp)));
			for (int j = 0; j < sz(dp); j++) {
				for (int k = 0; k < sz(dp[j]); k++) {
					st[k].update(j, dp[j][k]);
				}
			}
			for (int k = 0; k < sz(dp[0]); k++) {
				for (int l = 0; l < N; l++) {
					int ans = 0;
					{
						int l2 = 0, r = i-2 >= 0 ? sz(each[i-2]) - 1 : -1;
						while (l2 <= r) {
							int mid = (l2+r)/2;
							if (val[i-2][mid][2] <= val[i][l][0]) ans = mid, l2 = mid+1;
							else r = mid-1;
						}
					}
					{
						int L = k, R = val[i][l][0];
						ll cost = L <= R ? each[i-1][R][1] - each[i-1][L][1] : 0;
						ckmax(dp2[k][l], dp3[ans][k] + cost);
					}
					{
						if (ans + 1 < sz(dp)) {
							assert(i-2 >= 0);
							int l2 = ans+1, r2 = sz(dp)-1, ans2 = sz(dp);
							while (l2 <= r2) {
								int mid2 = (l2+r2)/2;
								if (val[i-2][mid2][2] >= k) ans2 = mid2, r2 = mid2-1;
								else l2 = mid2+1;
							}
							//for (int j = l2; j < ans2; j++) ckmax(dp2[k][l], dp[j][k]);
							if (l2 <= ans2 - 1) ckmax(dp2[k][l], st[k].query(l2, ans2 - 1));
							if (ans2 < sz(dp)) ckmax(dp2[k][l], dp4[ans2][k]);
						}
					}
				}
			}
			swap(dp, dp2);
		}
	}
	ll ans = -INF;
	for (int i = 0; i < sz(dp); i++) {
		for (int j = 0; j < sz(dp[i]); j++) {
			int L = j, R = val[n-2][i][2];
			ckmax(ans, dp[i][j] + (L <= R ? each[n-1][R][1] - each[n-1][L][1] : 0));
		}
	}
	return ans;
}

• Subtask 1: 3 / 3
• Subtask 2: 0 / 6
• Subtask 3: 9 / 9
• Subtask 4: 0 / 14
• Subtask 5: 0 / 21
• Subtask 6: 0 / 17
• Subtask 7: 0 / 14
• Subtask 8: 0 / 16