Submission #536141

#	Time	Username	Problem	Language	Result	Execution time	Memory
536141		browntoad	Collecting Stamps 3 (JOI20_ho_t3)	C++14	100 / 100	145 ms	135304 KiB

ho_t3

#include <bits/stdc++.h>
#pragma GCC optimize ("Ofast", "unroll-loops")
using namespace std;
#define ll long long
#define int ll
#define FOR(i,a,b) for (int i = (a); i<(b); i++)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) FOR(i,1,n+1)
#define RREP(i,n) for (int i=(n)-1; i>=0; i--)
#define RREP1(i,n) for (int i=(n); i>=1; i--)
#define f first
#define s second
#define pb push_back
#define ALL(x) x.begin(),x.end()
#define SZ(x) (int)(x.size())
#define SQ(x) (x)*(x)
#define pii pair<int, int>
#define pdd pair<double ,double>
#define pcc pair<char, char> 
#define endl '\n'
//#define TOAD
#ifdef TOAD
#define bug(x) cerr<<__LINE__<<": "<<#x<<" is "<<x<<endl
#define IOS()
#else
#define bug(...)
#define IOS() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#endif
 
const ll inf = 1ll<<60;
const int iinf=2147483647;
const ll mod = 1e9+7;
const ll maxn=205;
const double PI=acos(-1);
 
ll pw(ll x, ll p, ll m=mod){
    ll ret=1;
    while (p>0){
        if (p&1){
            ret*=x;
            ret%=m;
        }
        x*=x;
        x%=m;
        p>>=1;
    }
    return ret;
}
 
ll inv(ll a, ll m=mod){
    return pw(a,m-2);
}
 
//=======================================================================================

int dp[maxn][maxn][maxn][2];
signed main (){
    IOS();
    int n, L; cin>>n>>L;
    vector<int> t(n+2), vc(n+2);
    REP1(i,n){
        cin>>t[i];
    }
    t[n+1]=L;
    REP1(i,n){
        cin>>vc[i];
    }
    REP(i,maxn){
        REP(j, maxn){
            REP(k, maxn){
                REP(l, 2){
                    dp[i][j][k][l]=inf;
                }
            }
        }
    }
    dp[0][n+1][0][0]=dp[0][n+1][0][1]=0;

    REP(i,n+1){
        for (int j=n+1; j>i; j--){
            if (i==0&&j==n+1) continue;
            if (i>0) {
                REP(k,maxn){
                    dp[i][j][k][0]=min({dp[i][j][k][0], dp[i-1][j][k][0]+t[i]-t[i-1], dp[i-1][j][k][1]+(t[i]+L-t[j])});
                    if (k>0){
                        int a=dp[i-1][j][k-1][0]+t[i]-t[i-1];
                        int b=dp[i-1][j][k-1][1]+(t[i]+L-t[j]);
                        if (a<=vc[i]) dp[i][j][k][0]=min(dp[i][j][k][0], a);
                        if (b<=vc[i]) dp[i][j][k][0]=min(dp[i][j][k][0], b);
                    }
                }
            }
            if (j<=n) {
                REP(k,maxn){
                    dp[i][j][k][1]=min({dp[i][j][k][1], dp[i][j+1][k][1]+t[j+1]-t[j], dp[i][j+1][k][0]+(t[i]+L-t[j])});
                    if (k>0){
                        int a=dp[i][j+1][k-1][1]+t[j+1]-t[j];
                        int b=dp[i][j+1][k-1][0]+(t[i]+L-t[j]);
                        if (a<=vc[j]) dp[i][j][k][1]=min(dp[i][j][k][1], a);
                        if (b<=vc[j]) dp[i][j][k][1]=min(dp[i][j][k][1], b);
                    }
                }
            }
        }
    }
    //cout<<dp[0][3][1][1]<<endl;
    //cout<<dp[0][4][1][1]<<endl;
    //cout<<dp[0][4][0][1]<<endl;
    int ans=0;
    REP(k,maxn){
        REP(i,n+1){
            FOR(j,i+1,n+2){
                REP(l,2){
                    if (dp[i][j][k][l]!=inf) ans=k;
                }
            }
        }
    }
    cout<<ans<<endl;
}   

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