제출 #958891

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
958891		MinaRagy06	Shortcut (IOI16_shortcut)	C++17	31 / 100	2045 ms	94812 KiB

shortcut

#include <bits/stdc++.h>
#include "shortcut.h"
#ifdef MINA
#include "grader.cpp"
#endif
using namespace std;
#define ll long long

const int N = 3005;
ll v1[N][N], v2[N][N], v3[N][N], v4[N][N], mn1[N][N], mn2[N][N], mn3[N][N];
ll find_shortcut(int n, vector<int> l, vector<int> a, int c) {
	reverse(l.begin(), l.end()), l.push_back(0), reverse(l.begin(), l.end()), l.push_back(0);
	reverse(a.begin(), a.end()), a.push_back(-1), reverse(a.begin(), a.end());
	ll prf[n + 2]{};
	for (int i = 2; i <= n; i++) {
		prf[i] = prf[i - 1] + l[i - 1];
	}
	ll prfmx[n + 2]{}, prfd[n + 2]{}, sufmx[n + 2]{}, sufd[n + 2]{};
	for (int i = 1; i <= n; i++) {
		prfmx[i] = max(prfmx[i - 1] + l[i - 1], (ll)a[i]);
		prfd[i] = max(prfd[i - 1], prfmx[i - 1] + l[i - 1] + a[i]);
	}
	for (int i = n; i >= 1; i--) {
		sufmx[i] = max(sufmx[i + 1] + l[i], (ll)a[i]);
		sufd[i] = max(sufd[i + 1], sufmx[i + 1] + l[i] + a[i]);
	}
	auto check = [&] (ll t) {
		for (int i = 1; i <= n; i++) {
			for (int j = i + 1; j <= n; j++) {
				if (prf[j] - prf[i] + a[i] + a[j] <= t) {
					v1[i][j] = v2[i][j] = v3[i][j] = v4[i][j] = 1e18;
					continue;
				}
				v1[i][j] = t + prf[j] - prf[i] - a[i] - a[j];
				v2[i][j] = t - prf[j] + prf[i] - a[i] - a[j];
				v3[i][j] = t + prf[i] + prf[j] - a[i] - a[j];
				v4[i][j] = t - prf[i] - prf[j] - a[i] - a[j];
			}
		}
		for (int x = 1; x <= n; x++) {
			for (int y = x + 1; y <= n; y++) {
				mn1[x][y] = mn2[x][y] = mn3[x][y] = 1e18;
			}
		}
		for (int x = 1; x <= n; x++) {
			ll mn = 1e18, mn_ = 1e18;
			for (int y = x + 1; y <= n; y++) {
				for (int k = y - 1; k >= x; k--) {
					mn = min(mn, v1[k][y]);
				}
				mn1[x][y] = mn;
				for (int k = x; k >= 1; k--) {
					mn_ = min(mn_, v3[k][y]);
				}
				mn3[x][y] = mn_;
			}
			mn = mn_ = 1e18;
			for (int y = n; y >= x + 1; y--) {
				for (int k = x; k >= 1; k--) {
					mn = min(mn, v2[k][y]);
				}
				mn2[x][y] = mn;
			}
		}
		for (int y = 2; y <= n; y++) {
			ll mn4 = 1e18;
			for (int x = y - 1; x >= 1; x--) {
				for (int k = y; k <= n; k++) {
					mn4 = min(mn4, v4[x][k]);
				}
				ll len = prf[y] - prf[x] + c;
				if (len <= mn1[x][y] && -(prf[y] - prf[x]) + c <= mn2[x][y] && len + 2 * prf[x] <= mn3[x][y] &&
					len - 2 * prf[y] <= mn4 && prfd[x] <= t && sufd[y] <= t) {
					return 1;
				}
			}
		}
// 		for (int x = 1; x <= n; x++) {
// 			for (int y = x + 1; y <= n; y++) {
// 				ll mn1 = 1e18, mn2 = 1e18, mn3 = 1e18, mn4 = 1e18;
// 				ll len = prf[y] - prf[x] + c;
// 				for (int i = 1; i <= n; i++) {
// 					for (int j = i + 1; j <= n; j++) {
// 						if (x <= i && j <= y) {
// 							mn1 = min(mn1, v1[i][j]);
// 						} else if (i <= x && y <= j) {
// 							mn2 = min(mn2, v2[i][j]);
// 						} else if (i <= x && x <= j && j <= y) {
// 							mn3 = min(mn3, v3[i][j]);
// 						} else if (x <= i && i <= y && y <= j) {
// 							mn4 = min(mn4, v4[i][j]);
// 						} else {
// 						}
// 					}
// 				}
// 				if (len <= mn1 && -(prf[y] - prf[x]) + c <= mn2 && len + 2 * prf[x] <= mn3 && len - 2 * prf[y] <= mn4
// 						&& prfd[x] <= t && sufd[y] <= t) {
// 					return 1;
// 				}
// 			}
// 		}
		return 0;
	};
	ll s = 0, e = 1ll << 42;
	while (s <= e) {
		ll md = (s + e) >> 1;
		if (check(md)) {
			e = md - 1;
		} else {
			s = md + 1;
		}
	}
	return s;
}

• Subtask 1: 9 / 9
• Subtask 2: 14 / 14
• Subtask 3: 8 / 8
• Subtask 4: 0 / 7
• Subtask 5: 0 / 33
• Subtask 6: 0 / 22
• Subtask 7: 0 / 4
• Subtask 8: 0 / 3