Submission #724305

#	Time	Username	Problem	Language	Result	Execution time	Memory
724305		Cookie	Collecting Stamps 3 (JOI20_ho_t3)	C++14	100 / 100	101 ms	132896 KiB

ho_t3

#include<bits/stdc++.h>
#include<fstream>
using namespace std;
ifstream fin("WINTER.inp");
ofstream fout("WINTER.out");
#define ll long long
#define pb push_back
#define forr(i, a, b) for(int i = a; i < b; i++)
#define dorr(i, a, b) for(int i = a; i >= b; i--)
#define ld long double
#define vt vector
#include<fstream>
#define fi first
#define se second
#define pll pair<ll, ll>
#define pii pair<int, int>
const ld PI = 3.14159265359;
const ll mod = 1e9 + 7;
const int mxn = 2e5 + 5, mxm = 1e5;
ll dp[2][205][205][205],n , L, x[205], t[205];
void ckmin(ll &a, ll b){
    if(a > b)a = b;
}
signed main(){
    
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin >> n >> L;
    forr(i, 1, n + 1)cin >> x[i];
    forr(i, 1, n + 1)cin >> t[i];
    x[n + 1] = L;
    forr(i, 0, n + 2){
        forr(j, 0, n + 2){
            forr(k, 0, n + 2){
                dp[0][i][j][k] = dp[1][i][j][k] = 1e16;
            }
        }
    }
    dp[0][0][n + 1][0] = dp[1][0][n + 1][0] = 0;
   
    for(int l = 0; l < n; l++){
        for(int r = n + 1; r >= l + 2; r--){
            for(int take = 0; take <= n; take++){
                ll tt = min(dp[0][l][r][take] + abs(x[l + 1] - x[l]), dp[1][l][r][take] + L - abs(x[r] - x[l + 1]));
                ckmin(dp[0][l + 1][r][take + (tt <= t[l + 1])], tt);
                
                
                tt = min(dp[0][l][r][take] + L - abs(x[l] - x[r - 1]), dp[1][l][r][take] + abs(x[r] - x[r - 1]));
                ckmin(dp[1][l][r - 1][take + (tt <= t[r - 1])], tt);
                
            }
        }
    }
    
    int ans = 0;
    for(int i = 0; i <= n + 1; i++){
        for(int j = 0; j <= n + 1; j++){
            for(int k = 0; k <= n; k++){
                if(dp[0][i][j][k] < 1e16 || dp[1][i][j][k] < 1e16){
                    ans = max(ans, k);
                }
            }
        }
    }
    cout << ans;
    return(0);
}

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