Submission #1238356

#	Time	Username	Problem	Language	Result	Execution time	Memory
1238356		M_SH_O	Collecting Stamps 3 (JOI20_ho_t3)	C++20	100 / 100	114 ms	127560 KiB

ho_t3

/*#pragma GCC optimize("O3")
#pragma GCC optimization("Ofast,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")*/
#include <bits/stdc++.h>
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>*/

#define ll long long
#define ll1 long long
#define ull unsigned long long
#define dou long double
#define str string
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll, ll>
#define vpll vector<pll>
#define vbool vector<bool>
#define vstr vector<str>
#define vvll vector<vll>
#define pb push_back
#define pf push_front
#define endl "\n"
#define fr first
#define se second
// #define sortcmp(a) sort(a.begin(), a.end(), cmp)
#define sort(a) sort(a.begin(), a.end())
#define reverse(a) reverse(a.begin(), a.end())
#define speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define INF 1000000000000000007
#define ordered_set tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update>

using namespace std;
//using namespace __gnu_pbds;

mt19937 rng(time(0));
ll randll(ll l, ll r) {
    return uniform_int_distribution<ll>(l, r)(rng);
}

ll dp[201][201][201][2];

int main(){
    speed;
    srand(time(0));

    ll n, m;
    cin >> n >> m;
    vpll a(n);
    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] = INF;
            }
        }
    }

    for (int i = 0; i < n; i ++) {
        cin >> a[i].fr;
    }
    for (int i = 0; i < n; i ++) {
        cin >> a[i].se;
    }

    vpll b(n);
    b = a;
    reverse(b);
    for (int i = 0; i < n; i ++) {
        b[i].fr = m-b[i].fr;
    }


    dp[0][1][0][1] = a[0].fr;
    if (a[0].se >= a[0].fr) dp[0][1][1][1] = a[0].fr;
    dp[1][0][0][0] = b[0].fr;
    if (b[0].se >= b[0].fr) dp[1][0][1][0] = b[0].fr;

    for (int len = 1; len < n; len ++) {
        for (int i = 0; i <= len; i ++) {
            for (int j = 0; j <= len; j ++) {
                for (int q = 0; q < 2; q ++) {
                    if (dp[i][len-i][j][q] == INF) continue;
                    if (q == 0) {
                        ll t = dp[i][len-i][j][q] + b[i].fr-b[i-1].fr;
                        //if (len == 1 && i == 1) cout << dp[i][len-i][j][q] << ' ' << t << ' ' << b[i].fr << ' ' << b[i+1].fr << endl;
                        if (b[i].se >= t) dp[i+1][len-i][j+1][0] = min(dp[i+1][len-i][j+1][0], t);
                        else dp[i+1][len-i][j][0] = min(dp[i+1][len-i][j][0], t);

                        t = dp[i][len-i][j][q] + b[i-1].fr+a[len-i].fr;

                        if (a[len-i].se >= t) dp[i][len-i+1][j+1][1] = min(dp[i][len-i+1][j+1][1], t);
                        else dp[i][len-i+1][j][1] = min(dp[i][len-i+1][j][1], t);
                    }
                    else{
                        ll t = dp[i][len-i][j][q] + a[len-i-1].fr+b[i].fr;

                        if (b[i].se >= t) dp[i+1][len-i][j+1][0] = min(dp[i+1][len-i][j+1][0], t);
                        else dp[i+1][len-i][j][0] = min(dp[i+1][len-i][j][0], t);

                        t = dp[i][len-i][j][q] + a[len-i].fr-a[len-i-1].fr;

                        if (a[len-i].se >= t) dp[i][len-i+1][j+1][1] = min(dp[i][len-i+1][j+1][1], t);
                        else dp[i][len-i+1][j][1] = min(dp[i][len-i+1][j][1], t);
                    }
                }
            }
        }
    }
   // cout << dp[0][1][0][1] << endl;

    ll maxl = 0;
    for (ll i = 1; i <= n; i ++) {
        for (int j = 0; j <= n; j ++) {
            for (int k = 0; k <= n; k ++) {
                if (dp[j][k][i][0] != INF) {
                    maxl = i;
                    break;
                }
                if (dp[j][k][i][1] != INF) {
                    maxl = i;
                    break;
                }
            }
            if (maxl == i) break;
        }
    }

    cout << maxl << endl;
}

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