Submission #1022559

#	Time	Username	Problem	Language	Result	Execution time	Memory
1022559		RaresFelix	Shortcut (IOI16_shortcut)	C++17	71 / 100	2818 ms	524288 KiB

shortcut

#include "shortcut.h"
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2")

using namespace std;

using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;

const ll INF = 1e18;

const int MN = 3000;
ll DP1[MN][MN], DP2[MN][MN], DP3[MN][MN], DP4[MN][MN];

static bool posib(int n, vll alpha, vll beta, vll s, ll l, ll c) {
    vll gamma(n, 0);
    for(int i = 0; i < n; ++i)
        gamma[i] = l - alpha[i];
    for(int i = 0; i < n; ++i)
        DP1[0][i] = DP2[0][i] = DP3[0][i] = DP4[0][i] = -INF;
    for(int d = 1; d < n; ++d) {
        for(int i = 0; i < n; ++i) {
            int j = i + d;
            if(j >= n) break;
            if(beta[j] > gamma[i]) {
                DP1[j - i][i] = alpha[i] + alpha[j];
                DP2[j - i][i] = beta[j] + alpha[i];
                DP3[j - i][i] = beta[i] + alpha[j];
                DP4[j - i][i] = beta[i] + beta[j];
            } else 
                DP1[j - i][i] = DP2[j - i][i] = DP3[j - i][i] = DP4[j - i][i] = -INF;
        }
    }
    for(int len = 1; len < n; ++len) 
        for(int i = 0; i < n; ++i) {
            int j = i + len;
            if(j >= n) break;
            if(j) {
                DP1[j - i][i] = max(DP1[len][i], DP1[len - 1][i]);
                DP3[j - i][i] = max(DP3[len][i], DP3[len - 1][i]);
            }
            if(i + 1 < n && len) {
                DP3[len][i] = max(DP3[len][i], DP3[len - 1][i + 1]);
                DP4[len][i] = max(DP4[len][i], DP4[len - 1][i + 1]);
            }
        }
    for(int len = n - 1; len >= 0; --len)
        for(int i = 0; i < n; ++i) {
            int j = i + len;
            if(j >= n) break;
            if(i && len + 1 < n) {
                DP1[len][i] = max(DP1[len][i], DP1[len + 1][i - 1]);
                DP2[len][i] = max(DP2[len][i], DP2[len + 1][i - 1]);
            }
            if(j + 1 < n && len + 1 < n) {
                DP2[len][i] = max(DP2[len][i], DP2[len + 1][i]);
                DP4[len][i] = max(DP4[len][i], DP4[len + 1][i]);
            }
        }
    for(int d = 0; d < n; ++d) {
        for(int a = 0; a < n; ++a) {
            int b = d + a;
            if(b >= n) break;
            bool ok = true;
            ok &= DP1[b - a][a] + s[a] + s[b] + c <= l;
            ok &= DP2[b - a][a] + s[a] - s[b] + c <= l;
            ok &= DP3[b - a][a] - s[a] + s[b] + c <= l;
            ok &= DP4[b - a][a] - s[a] - s[b] + c <= l;
            if(ok) return true;
        }
    }
    return false;
}

ll find_shortcut(int n, vi l, vi d, int c) {
    vll s(n, 0), alpha(n, 0), beta(n, 0);

    for(int i = 1; i < n; ++i) {
        s[i] = s[i - 1] + l[i - 1];
    }
    for(int i = 0; i < n; ++i) {
        alpha[i] = d[i] - s[i];
        beta[i] = d[i] + s[i];
    }
    ll st = 0, dr = (n + 2) * ll(1e9), mij;
    while(st < dr) {
        mij = (st + dr) / 2;
        if(posib(n, alpha, beta, s, mij, c))
            dr = mij;
        else st = mij + 1;
    }

    return st;
}

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