Submission #206498

#	Time	Username	Problem	Language	Result	Execution time	Memory
206498		opukittpceno_hhr	Semiexpress (JOI17_semiexpress)	C++17	48 / 100	1092 ms	70904 KiB

semiexpress

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <complex>
#include <random>

using namespace std;

#define int long long

const int N = 3001;
const int INF = 1e18 + 239;

int dp[N][N];

int n, k;
int lc, ex, sm;
int t;

vector<int> proc(int sd, int len) {
	if (sd >= t) {
		return {0};
	}
	int wd = sd;
	int delta = t - sd;
	int can = delta / lc;
	vector<int> ret = {can};
	sd += (can) * sm;
	if (ret.back() < len) {
		for (int i = 0; i < k; i++) {
			sd += sm;
			if (sd > t) {
				break;
			}
			can = (t - sd) / lc;
			ret.push_back(ret.back() + can + 1);
			sd += can * sm;
			if (ret.back() > len) {
				ret.back() = len;
				break;
			}
		}
	} else {
		ret.back() = len;
	}
	for (auto &x : ret) {
		if (x < len && wd + len * ex <= t) {
			x++;
		}
	}
	return ret;
}


signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			dp[i][j] = -INF;
		}
	}

	int m;
	cin >> n >> m >> k;
	k -= m;
	cin >> lc >> ex >> sm;
	cin >> t;
	vector<int> s(m);
	for (auto &x : s) {
		cin >> x;
		x--;
	}
	dp[0][0] = 0;
	vector<pair<int, int>> v;
	for (int i = 1; i < m; i++) {
		int len = s[i] - s[i - 1];
		int sd = s[i - 1] * ex;
		auto nw = proc(sd, len);
		if (nw.empty()) continue;
		vector<int> dl = {nw[0]};
		for (int j = 1; j < (int)dl.size(); j++) {
			dl.push_back(nw[j] - nw[j - 1]);
		}
		{
			reverse(dl.begin(), dl.end());
			assert(is_sorted(dl.begin(), dl.end()));
		}
		for (int j = 0; j < N; j++) {
			if (dp[i - 1][j] == -INF) continue;
			for (int tk = min((int)nw.size(), N - j) - 1; tk >= 0; tk--) {
				dp[i][j + tk] = max(dp[i][j + tk], dp[i - 1][j] + nw[tk]);
			}
		}
	}
	int ans = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j <= k; j++) {
			ans = max(ans, dp[i][j]);
		}
	}
	cout << ans << endl;
}

• Subtask 1: 18 / 18
• Subtask 2: 30 / 30
• Subtask 3: 0 / 52