제출 #987412

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
987412		GrindMachine	Two Antennas (JOI19_antennas)	C++17	100 / 100	1085 ms	58424 KiB

antennas

#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:
https://codeforces.com/blog/entry/66022?#comment-500622
https://mzhang2021.github.io/cp-blog/historic-segtree/ (historic sums segment tree)

*/

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

template<typename T>
struct lazysegtree {
    /*=======================================================*/

    struct data {
        ll a,b;
    };

    struct lazy {
        ll sum,mx_pref;
    };

    data d_neutral = {-inf2,-inf2};
    lazy l_neutral = {0,0};

    void merge(data &curr, data &left, data &right) {
        curr.a = max(left.a,right.a);
        curr.b = max(left.b,right.b);
    }

    void create(int x, int lx, int rx, T v) {

    }

    void modify(int x, int lx, int rx, T v) {
        if(v.ff == 1){
            lz[x].sum = v.ss;
            lz[x].mx_pref = max(v.ss,0ll);
        }
        else{
            tr[x].a = v.ss;
            amax(tr[x].b,v.ss);
        }
    }

    void propagate(int x, int lx, int rx) {
        auto [sum,mx_pref] = lz[x];
        amax(tr[x].b,tr[x].a+mx_pref);
        tr[x].a += sum;

        if(rx-lx > 1){
            lz[2*x+1].mx_pref = max(lz[2*x+1].mx_pref,lz[2*x+1].sum+mx_pref);
            lz[2*x+1].sum += sum;

            lz[2*x+2].mx_pref = max(lz[2*x+2].mx_pref,lz[2*x+2].sum+mx_pref);
            lz[2*x+2].sum += sum;
        }

        lz[x] = l_neutral;
    }

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

    int siz = 1;
    vector<data> tr;
    vector<lazy> lz;

    lazysegtree() {

    }

    lazysegtree(int n) {
        while (siz < n) siz *= 2;
        tr.assign(2 * siz, d_neutral);
        lz.assign(2 * siz, l_neutral);
    }

    void build(vector<T> &a, int n, int x, int lx, int rx) {
        if (rx - lx == 1) {
            if (lx < n) {
                create(x, lx, rx, a[lx]);
            }

            return;
        }

        int mid = (lx + rx) / 2;

        build(a, n, 2 * x + 1, lx, mid);
        build(a, n, 2 * x + 2, mid, rx);

        merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
    }

    void build(vector<T> &a, int n) {
        build(a, n, 0, 0, siz);
    }

    void rupd(int l, int r, T v, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return;
        if (lx >= l and rx <= r) {
            modify(x, lx, rx, v);
            propagate(x, lx, rx);
            return;
        }

        int mid = (lx + rx) / 2;

        rupd(l, r, v, 2 * x + 1, lx, mid);
        rupd(l, r, v, 2 * x + 2, mid, rx);

        merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
    }

    void rupd(int l, int r, T v) {
        rupd(l, r + 1, v, 0, 0, siz);
    }

    data query(int l, int r, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return d_neutral;
        if (lx >= l and rx <= r) return tr[x];

        int mid = (lx + rx) / 2;

        data curr;
        data left = query(l, r, 2 * x + 1, lx, mid);
        data right = query(l, r, 2 * x + 2, mid, rx);

        merge(curr, left, right);
        return curr;
    }

    data query(int l, int r) {
        return query(l, r + 1, 0, 0, siz);
    }
};

void solve(int test_case)
{
    ll n; cin >> n;
    vector<ll> a(n+5);
    vector<pll> b(n+5);
    rep1(i,n) cin >> a[i] >> b[i].ff >> b[i].ss;
    ll q; cin >> q;
    vector<pll> queries(q+5);
    rep1(i,q) cin >> queries[i].ff >> queries[i].ss;
        
    vector<ll> ans(q+5,-inf2);

    auto go = [&](){
        vector<ll> enter[n+5], leave[n+5];
        vector<pll> queries_here[n+5];
        lazysegtree<pll> st(n+5);

        rep1(i,n){
            auto [l,r] = b[i];
            if(i+l <= n){
                enter[i+l].pb(i);
            }
            if(i+r+1 <= n){
                leave[i+r+1].pb(i);
            }
        }

        rep1(i,q){
            auto [l,r] = queries[i];
            queries_here[r].pb({l,i});
        }

        rep1(i,n){
            trav(j,enter[i]){
                st.rupd(j,j,{2,-a[j]});
            }
            trav(j,leave[i]){
                st.rupd(j,j,{2,-inf2});
            }

            auto [l,r] = b[i];
            l = i-l, r = i-r;
            if(l > r) swap(l,r);
            amax(l,1ll);

            st.rupd(l,r,{1,a[i]});
            st.rupd(l,r,{1,-a[i]});

            for(auto [l,id] : queries_here[i]){
                amax(ans[id],st.query(l,i).b);
            }
        }
    };

    go();

    reverse(a.begin()+1,a.begin()+n+1);
    reverse(b.begin()+1,b.begin()+n+1);
    rep1(i,q){
        auto &[l,r] = queries[i];
        l = n-l+1;
        r = n-r+1;
        swap(l,r);
    }

    go();

    rep1(i,q){
        if(ans[i] < 0){
            ans[i] = -1;
        }
        cout << ans[i] << endl;
    }
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 2 / 2
• Subtask 2: 11 / 11
• Subtask 3: 22 / 22
• Subtask 4: 65 / 65