Submission #545306

#	Time	Username	Problem	Language	Result	Execution time	Memory
545306		balbit	Visiting Singapore (NOI20_visitingsingapore)	C++14	12 / 100	121 ms	480 KiB

VisitingSingapore

#include <bits/stdc++.h>
#define int ll
using namespace std;
#define ll long long
#define y1 zck_is_king
#define pii pair<ll, ll>
#define ull unsigned ll
#define f first
#define s second
#define ALL(x) x.begin(),x.end()
#define SZ(x) (int)x.size()
#define SQ(x) (x)*(x)
#define MN(a,b) a = min(a,(__typeof__(a))(b))
#define MX(a,b) a = max(a,(__typeof__(a))(b))
#define pb push_back
#define REP(i,n) for (int i = 0; i<n; ++i)
#define RREP(i,n) for (int i = n-1; i>=0; --i)
#define REP1(i,n) for (int i = 1; i<=n; ++i)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#ifdef BALBIT
#define IOS()
#define bug(...) fprintf(stderr,"#%d (%s) = ",__LINE__,#__VA_ARGS__),_do(__VA_ARGS__);
template<typename T> void _do(T &&x){cerr<<x<<endl;}
template<typename T, typename ...S> void _do(T &&x, S &&...y){cerr<<x<<", ";_do(y...);}
#else
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
#define bug(...)
#endif

const int iinf = 1e9+10;
const ll inf = 1ll<<60;
const ll mod = 1e9+7 ;


void GG(){cout<<"0\n"; exit(0);}

ll mpow(ll a, ll n, ll mo = mod){ // a^n % mod
    ll re=1;
    while (n>0){
        if (n&1) re = re*a %mo;
        a = a*a %mo;
        n>>=1;
    }
    return re;
}

ll inv (ll b, ll mo = mod){
    if (b==1) return b;
    return (mo-mo/b) * inv(mo%b,mo) % mo;
}

const int maxn = 5005+5;

ll V[maxn], S[maxn], T[maxn];
ll dp[2][maxn];

signed main(){
    IOS();
    int K,n,m,A,B;
    cin>>K>>n>>m>>A>>B;
    assert(A == 0 && B == 0);

    REP1(i,K) cin>>V[i];
    REP1(i,n) cin>>S[i];
    REP1(i,m) cin>>T[i];

//    dp[0][1][0][1] = 0;-
    // calculate cost of not even visiting singapore
//    ll re = (A + B*m);
    ll re = 0;
    REP1(i,n) {
        int I = i%2;
        REP1(j,m) {
            dp[I][j] = -inf;
            dp[I][j] = max(dp[I^1][j], dp[I][j-1]);
            if (S[i] == T[j]) {
                MX(dp[I][j], dp[I^1][j-1] + V[S[i]]);
            }
            MX(re, dp[I][j]);
        }
    }
    cout<<re<<endl;
}

• Subtask 1: 0 / 4
• Subtask 2: 0 / 6
• Subtask 3: 12 / 12
• Subtask 4: 0 / 7
• Subtask 5: 0 / 8
• Subtask 6: 0 / 13
• Subtask 7: 0 / 50