Submission #747413

#	Time	Username	Problem	Language	Result	Execution time	Memory
747413		GrindMachine	Hedgehog Daniyar and Algorithms (IZhO19_sortbooks)	C++17	100 / 100	991 ms	110132 KiB

sortbooks

// Om Namah Shivaya

#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) 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(x) 42
#endif

/*

guy j has to cross guy i (i < j) if:
a[j] < a[i]

for all guys that must cross each other, find if a[i]+a[j] <= k

rewriting the problem:
for given (l,r), find the max val of a[i] + a[j], s.t:
i < j and a[i] > a[j]

i.e find max sum inversion pair in range

we can try to use sweepline

sweep over r, maintaining the max ans for all l
so when answering a query, we can just find the max sum over all l in range [l,r] and check if it's > k

what changes when we move from r-1 to r

a[r] forms an inversion pair only with a[i] > a[r], i < r
let the indices of all guys < r with val > a[r] in descending order of indices be:
i1 > i2 > ... > ik

(i1,r) forms an inversion pair, so we can upd the val at i1 to a[i1] + a[r]

(i2,r) also forms an inversion pair

there are 2 cases to consider for i2:
1) a[i2] <= a[i1]: (i1,r) gives better result than (i2,r), so no need to update
2) a[i2] > a[i1]: (i2,i1) gives better result than (i2,r), so no need to update

similar arguments apply for i3,i4,...,ik

so we just have to upd the val at i1
i1 = next greater value to the left of a[r]
which can be found using a stack

updates and range max queries can be handled with a segtree

*/

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 segtree {
    // https://codeforces.com/blog/entry/18051

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

    struct data {
        ll a;
    };

    data neutral = {-inf2};

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

    void create(int i, T v) {

    }

    void modify(int i, T v) {
        amax(tr[i].a, 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);
    }
};

void solve(int test_case)
{
    ll n,m; cin >> n >> m;
    vector<ll> a(n+5);
    rep1(i,n) cin >> a[i];

    stack<ll> st;
    vector<ll> nge(n+5,-1);

    rev(i,n,1){
        while(!st.empty() and a[i] > a[st.top()]){
            nge[st.top()] = i;
            st.pop();
        }
        st.push(i);
    }

    vector<array<ll,3>> queries[n+5];

    rep1(i,m){
        ll l,r,k; cin >> l >> r >> k;
        queries[r].pb({l,k,i});
    }

    segtree<ll> seg(n+5);
    vector<ll> ans(m+5);

    rep1(r,n){
        ll j = nge[r];
        if(j != -1){
            seg.pupd(j, a[r] + a[j]);            
        }

        for(auto [l,k,id] : queries[r]){
            ll mx = seg.query(l,r).a;
            if(mx > k) ans[id] = 0;
            else ans[id] = 1;
        }
    }

    rep1(i,m) cout << ans[i] << endl;
}

int main()
{
    fastio;

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

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

    return 0;
}

• Subtask 1: 8 / 8
• Subtask 2: 9 / 9
• Subtask 3: 13 / 13
• Subtask 4: 17 / 17
• Subtask 5: 30 / 30
• Subtask 6: 23 / 23