제출 #1058634

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1058634		Jarif_Rahman	Two Antennas (JOI19_antennas)	C++17	22 / 100	141 ms	26964 KiB

antennas

#include <bits/stdc++.h>
#define pb push_back
#define f first
#define sc second
using namespace std;
typedef long long int ll;
typedef string str;

const int inf = 1000000005;

struct segtree{
    struct Node{
        int mx, mn;
        Node(){
             mx = -inf, mn = inf;
        }
        Node(int x){
            mx = x, mn = x;
        }
        Node operator+(Node a){
            Node rt;
            rt.mx = max(mx, a.mx);
            rt.mn = min(mn, a.mn);
            return rt;
        }
    };
    int k;
    vector<Node> v;
    segtree(int n){
        k = 1;
        while(k < n) k*=2;
        v.resize(2*k, Node());
    }
    void update(int in, int x){
        in+=k;
        if(x == -1) v[in] = Node();
        else v[in] = Node(x);
        in/=2;
        while(in > 0){
            v[in] = v[2*in]+v[2*in+1];
            in/=2;
        }
    }
    Node get(int l, int r, int nd, int a, int b){
        if(b < l || a > r) return Node();
        if(a >= l && b <= r) return v[nd];
        int c = (a+b)/2;
        return get(l, r, 2*nd, a, c) + get(l, r, 2*nd+1, c+1, b);
    }
    Node get(int l, int r){
        return get(l, r, 1, 0, k-1);
    }
};

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int n; cin >> n;
    vector<int> H(n), A(n), B(n);
    for(int i = 0; i < n; i++){
        cin >> H[i] >> A[i] >> B[i];
    }

    segtree S(n);
    for(int i = 0; i < n; i++){
        if(i-B[i] < 0 && i-A[i] >= 0) S.update(i, H[i]);
    }

    vector<vector<int>> st(n), ed(n);
    for(int i = 0; i < n; i++){
        if(i-B[i] >= 0) st[i-B[i]].push_back(i);
        if(i-A[i] >= 0) ed[i-A[i]].push_back(i);
        if(i+A[i] < n) st[i+A[i]].push_back(i);
        if(i+B[i] < n) ed[i+B[i]].push_back(i);
    }

    int ans = -1;
    for(int i = 0; i < n; i++){
        for(int j: st[i]) S.update(j, H[j]);
        segtree::Node nd;
        if(i-A[i] >= 0) nd = nd+S.get(max(0, i-B[i]), i-A[i]);
        if(i+A[i] < n) nd = nd+S.get(i+A[i], min(i+B[i], n-1));
        if(nd.mx > -inf){
            ans = max(ans, max(abs(H[i]-nd.mx), abs(H[i]-nd.mn)));
        }
        for(int j: ed[i]) S.update(j, -1);
    }

    cout << ans << "\n";
}

• Subtask 1: 0 / 2
• Subtask 2: 0 / 11
• Subtask 3: 22 / 22
• Subtask 4: 0 / 65