Submission #1016538

#	Time	Username	Problem	Language	Result	Execution time	Memory
1016538		GrindMachine	Homecoming (BOI18_homecoming)	C++17	31 / 100	1058 ms	28220 KiB

homecoming

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "homecoming.h"

ll solve(int n, int k, int *A, int *B){
    vector<ll> a(n), b(n);
    rep(i,n) a[i] = A[i], b[i] = B[i];
    ll initial = 0;
    rep(i,n) initial += a[i]-b[i];
    ll ans = 0;

    rep(cyc_shift,n){
        if(cyc_shift){
            rotate(a.begin(),a.begin()+1,a.end());
            rotate(b.begin(),b.begin()+1,b.end());
        }

        vector<ll> p(n);
        rep(i,n){
            p[i] = a[i]+(i?p[i-1]:0);
        }

        vector<ll> dp(n,-inf2);
        vector<ll> pmx1(n,-inf2), pmx2(n,-inf2);

        for(int i = k; i < n; ++i){
            dp[i] = b[i]-(p[i]-p[i-k]);
            ll val1 = pmx1[i-1]+b[i]-p[i];
            ll val2 = pmx2[i-k]+b[i]-(p[i]-p[i-k]);
            dp[i] = max({dp[i],val1,val2});
            pmx1[i] = max(pmx1[i-1],dp[i]+p[i]);
            pmx2[i] = max(pmx2[i-1],dp[i]);
        }

        amax(ans,*max_element(all(dp)));

        // trav(x,a) cout << x << " ";
        // cout << endl;
        
        // trav(x,b) cout << x << " ";
        // cout << endl;
    }

    ans += initial;
    amax(ans,0ll);
    return ans;
}

• Subtask 1: 13 / 13
• Subtask 2: 18 / 18
• Subtask 3: 0 / 31
• Subtask 4: 0 / 38