제출 #1065811

#제출 시각아이디문제언어결과실행 시간메모리
1065811phoenixCollecting Stamps 3 (JOI20_ho_t3)C++17
100 / 100
189 ms141140 KiB
#include <bits/stdc++.h>

using namespace std;

#define cerr if (false) cout
using ll = long long;

const int N = 220;
const ll INF = 1e18;

int n, L;
int x[N];
int t[N];

ll dp[N][N][N][2];

int main() {
    // ios::sync_with_stdio(0);
    // cin.tie(0);
    cin >> n >> L;
    for (int i = 1; i <= n; i++)
        cin >> x[i];
    for (int i = 1; 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] = INF,
                dp[i][j][k][1] = INF;
    
    dp[0][0][0][0] = 0;
    dp[0][0][0][1] = 0;
    for (int k = 0; k < n; k++) {
        for (int s = 0; s < n; s++) {
            for (int l = 0; l <= s; l++) {
                int r = (l + n + 1 - s) % (n + 1);
                cerr << l << ' ' << r << endl;
                ll time, k_new;
                time = dp[l][r][k][0] + x[l + 1] - x[l];
                k_new = k + (time <= t[l + 1]);
                dp[l + 1][r][k_new][0] = min(dp[l + 1][r][k_new][0], time);
                cerr << "one\n";

                time = dp[l][r][k][1] + x[l + 1] + (L - x[r]) % L;
                k_new = k + (time <= t[l + 1]);
                dp[l + 1][r][k_new][0] = min(dp[l + 1][r][k_new][0], time);
                cerr << "two\n";
                
                int r1 = (r + n) % (n + 1);
                
                time = dp[l][r][k][0] + x[l] + L - x[r1];
                k_new = k + (time <= t[r1]);
                dp[l][r1][k_new][1] = min(dp[l][r1][k_new][1], time);
                cerr << "three\n";

                time = dp[l][r][k][1] + (x[r] - x[r1] + L) % L;
                k_new = k + (time <= t[r1]);
                dp[l][r1][k_new][1] = min(dp[l][r1][k_new][1], time);
                cerr << "four\n";
            }
        }
    }

    for (int k = n; k >= 0; k--) {
        for (int i = 0; i <= n; i++) {
            if (dp[i][(i + 1) % (n + 1)][k][0] != INF) {
                cout << k;
                return 0;
            }   
            if (dp[i][(i + 1) % (n + 1)][k][1] != INF) {
                cout << k;
                return 0;
            }   
        }
    }
}
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