제출 #1019082

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1019082		GrindMachine	Homecoming (BOI18_homecoming)	C++17	0 / 100	1012 ms	161228 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) {
        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;
    }

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

    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){
    // all subjects win (so all books taken)
    ll all_win = 0;
    rep(i,n) all_win += A[i]-B[i];

    ll ans = max(all_win,0ll);

    segtree<pll> st(n+5);
    ll s = 0, cost2 = 0;
    rep(i,k) s += B[i];

    rep1(i,n){
        if(i != 1){
            s -= B[(i-1-1)%n];
            s += B[(i+k-1-1)%n];
        }

        ll cost1 = A[(i-1)%n]-s;
        cost2 += A[(i-1)%n]-B[(i+k-1-1)%n];
        cost1 -= cost2;
        st.pupd(i,{cost1,cost2});
    }

    // at least 1 subject doesnt win (so at least 1 book not taken)
    rep1(f,n){
        // f = first book not taken
        ll l = f+1, r = f+n-k;
        if(l > r) conts;
        l--, r--;
        l %= n, r %= n;
        ll val = 0;

        if(l <= r){
            val = st.query(l+1,r+1).a[0][0];
        }
        else{
            auto d1 = st.query(l+1,n);
            auto d2 = st.query(1,r+1);
            val = st.merge(d1,d2).a[0][0];
        }

        amax(ans,val);
    }

    return ans;
}

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