Submission #535786

#	Time	Username	Problem	Language	Result	Execution time	Memory
535786		abc864197532	Collecting Stamps 3 (JOI20_ho_t3)	C++17	100 / 100	106 ms	67780 KiB

ho_t3

#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define mp make_pair
#define eb emplace_back
#define pb push_back
#define pii pair<int,int>
#define X first
#define Y second
#define all(x) x.begin(), x.end()
void abc() {cout << endl;}
template <typename T, typename ...U> void abc(T i, U ...j) {
	cout << i << ' ', abc(j...);
}
template <typename T> void printv(T l, T r) {
	for (; l != r; ++l)
		cout << *l << " \n"[l + 1 == r];
}
#ifdef Doludu
#define test(x...) abc("[" + string(#x) + "]", x)
#define owo freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#else
#define test(x...) void(0)
#define owo ios::sync_with_stdio(false), cin.tie(0)
#endif
const int N = 205;

int dp[N][N][N][2];

void upd(int &x, int y) {
	x = min(x, y);
}

int main () {
	owo;
	int n, l;
	cin >> n >> l;
	vector <int> x(n), T(n);
	for (int i = 0; i < n; ++i)
		cin >> x[i];
	for (int i = 0; i < n; ++i)
		cin >> T[i];
	for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) for (int k = 0; k < N; ++k)
		dp[i][j][k][0] = dp[i][j][k][1] = 1 << 30; 
	auto dis = [&](int x, int y) {
		int d = abs(y - x);
		return min(d, l - d);
	};
	{
		int t = min(x[0], l - x[0]);
		dp[1][0][t <= T[0]][0] = t;
	}
	{
		int t = min(x[n - 1], l - x[n - 1]);
		dp[0][1][t <= T[n - 1]][1] = t;
	}
	for (int len = 1; len < n; ++len) for (int l = 0; l <= len; ++l) for (int num = 0; num < n; ++num) {
		int r = len - l;
		// left to left
		if (l && dp[l][r][num][0] != 1 << 30) {
			int t = dp[l][r][num][0] + dis(x[l - 1], x[l]);
			upd(dp[l + 1][r][num + (t <= T[l])][0], t);
		}
		// left to right
		if (l && dp[l][r][num][0] != 1 << 30) {
			int t = dp[l][r][num][0] + dis(x[l - 1], x[n - r - 1]);
			upd(dp[l][r + 1][num + (t <= T[n - r - 1])][1], t);
		}
		// right to left
		if (r && dp[l][r][num][1] != 1 << 30) {
			int t = dp[l][r][num][1] + dis(x[n - r], x[l]);
			upd(dp[l + 1][r][num + (t <= T[l])][0], t);
		}
		// right to right
		if (r && dp[l][r][num][1] != 1 << 30) {
			int t = dp[l][r][num][1] + dis(x[n - r], x[n - r - 1]);
			upd(dp[l][r + 1][num + (t <= T[n - r - 1])][1], t);
		}
	}
	int ans = 0;
	for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) for (int k = 0; k < N; ++k) {
		if (dp[i][j][k][0] < 1 << 30 || dp[i][j][k][1] < 1 << 30)
			ans = max(ans, k);
	}
	cout << ans << endl;
}

• Subtask 1: 5 / 5
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 75 / 75