Submission #1019053

#	Time	Username	Problem	Language	Result	Execution time	Memory
1019053		GrindMachine	Homecoming (BOI18_homecoming)	C++17	31 / 100	213 ms	262144 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

/*

refs:
edi

*/

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"

template<typename T>
struct segtree {
    // https://codeforces.com/blog/entry/18051

    /*=======================================================*/

    struct data {
        ll a[2][2];
        data(){
            rep(i,2) rep(j,2) a[i][j] = -inf2;
        }
    };

    data neutral = data();

    data merge(data &left, data &right) {
        data curr;
        
        rep(a,2){
            rep(b,2){
                amax(curr.a[a][b],left.a[a][b]);
                amax(curr.a[a][b],right.a[a][b]);

                rep(c,2){
                    amax(curr.a[a][c],left.a[a][b]+right.a[b][c]);
                }
            }
        }

        return curr;
    }

    void create(int i, T v) {
        auto [x,y] = v;
        tr[i].a[0][0] = max(x+y,0ll);
        tr[i].a[0][1] = x;
        tr[i].a[1][0] = y;
        tr[i].a[1][1] = -inf2;
    }

    void modify(int i, T v) {

    }

    /*=======================================================*/

    int n;
    vector<data> tr;

    segtree() {

    }

    segtree(int siz) {
        init(siz);
    }

    void init(int siz) {
        n = siz;
        tr.assign(2 * n, neutral);
    }

    void build(vector<T> &a, int siz) {
        rep(i, siz) create(i + n, a[i]);
        rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    void pupd(int i, T v) {
        modify(i + n, v);
        for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    data query(int l, int r) {
        data resl = neutral, resr = neutral;

        for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
            if (l & 1) resl = merge(resl, tr[l++]);
            if (!(r & 1)) resr = merge(tr[r--], resr);
        }

        return merge(resl, resr);
    }
};

ll solve(int n, int k, int *A, int *B){
    vector<int> a(2*n+5), b(2*n+5);
    rep(i,2*n) a[i+1] = A[i%n], b[i+1] = B[i%n];

    vector<ll> pb(2*n+5);
    rep1(i,2*n) pb[i] = pb[i-1]+b[i];

    vector<ll> cost1(2*n+5), cost2(2*n+5);
    rep1(i,2*n){
        if(i+k-1 > 2*n) break;
        ll s = pb[i+k-1]-pb[i-1];
        cost1[i] = a[i]-s;
        cost2[i] = a[i]-b[i+k-1];
    }
    rep1(i,2*n) cost2[i] += cost2[i-1];
    rep1(i,2*n) cost1[i] -= cost2[i];

    // all subjects win (so all books taken)
    ll all_win = 0;
    rep1(i,n) all_win += a[i]-b[i];

    ll ans = max(all_win,0ll);

    segtree<pll> st(2*n+5);
    vector<pll> curr(2*n+5);
    rep1(i,2*n) curr[i] = {cost1[i],cost2[i]};
    st.build(curr,2*n+1);

    // at least 1 subject doesnt win (so at least 1 book not taken)
    rep1(f,n){
        // f = first book not taken
        ll dp[2];
        memset(dp,-0x3f,sizeof dp);
        dp[0] = 0;
        ll val = st.query(f+1,f+n-k).a[0][0];
        amax(ans,val);

        // for(int i = f+1; i+k-1 < f+n; ++i){
        //     amax(dp[1],dp[0]+cost1[i]);
        //     amax(dp[0],dp[1]+cost2[i]);
        // }

        // amax(ans,dp[0]);
    }

    // ll oans = all_win;

    // rep1(f,n){
    //     // f = first book not taken
    //     ll dp[2];
    //     memset(dp,-0x3f,sizeof dp);
    //     dp[0] = 0;
 
    //     for(int i = f; i+k-1 < f+n; ++i){
    //         amax(dp[1],dp[0]+cost1[i]);
    //         amax(dp[0],dp[1]+cost2[i]);
    //     }
 
    //     amax(oans,dp[0]);
    // }

    // assert(ans == oans);
    return ans;
}

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