Submission #61842

#	Time	Username	Problem	Language	Result	Execution time	Memory
61842		Benq	Homecoming (BOI18_homecoming)	C++11	100 / 100	289 ms	119488 KiB

homecoming

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

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 2000001;

#include "homecoming.h"

ll cb[2*MX], dp[MX][2][2];

ll get(int l, int r) { 
    // cout << "OH " << cb[r+1]-cb[l] << "\n";
    return cb[r+1]-cb[l]; 
}

long long int solve(int N, int K, int *A, int *B) {
    F0R(i,2*N) cb[i+1] = cb[i]+B[i%N];
    
    ll ans = 0;
    F0R(i,N) F0R(j,2) F0R(k,2) dp[i][j][k] = -INF;
    dp[0][0][0] = 0, dp[0][1][1] = A[0]-get(0,K-1);
    FOR(i,1,N) {
        F0R(j,2) {
            dp[i][0][j] = max(dp[i-1][0][j],dp[i-1][1][j]);
            if (j == 0 || i+K-1 < N) {
                dp[i][1][j] = max(dp[i-1][1][j]+A[i]-B[(i+K-1)%N],dp[i-1][0][j]+A[i]-get(i,i+K-1));
            } else {
                dp[i][1][j] = max(dp[i-1][1][j]+A[i],dp[i-1][0][j]+A[i]-get(i,N-1));
            }
            // cout << "OH " << i << " " << j << " " << dp[i][0][j] << " " << dp[i][1][j] << "\n";
        } 
    }
    F0R(j,2) F0R(k,2) ans = max(ans,dp[N-1][j][k]);
    return ans;
}

/*int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int A[3] = {40, 80, 100};
    int B[3] = {140, 0, 20};
    cout << solve(3,2,A,B);
}*/

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

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