제출 #655725

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
655725		benjaminkleyn	Foehn Phenomena (JOI17_foehn_phenomena)	C++17	100 / 100	285 ms	15616 KiB

foehn_phenomena

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int N, Q;
ll S, T, A[300001];

ll t[1000000];
void build(int v, int tl, int tr)
{
    if (tl == tr)
    {
        t[v] = A[tl];
        return;
    }
    int tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    t[v] = 0;
}

void add(int v, int tl, int tr, int l, int r, ll val)
{
    if (r < tl || tr < l)
        return;
    if (l <= tl && tr <= r)
    {
        t[v] += val;
        return;
    }
    int tm = (tl + tr) / 2;
    add(v * 2, tl, tm, l, r, val);
    add(v * 2 + 1, tm + 1, tr, l, r, val);
}

ll query(int v, int tl, int tr, int idx)
{
    if (tl == tr)
        return t[v];
    int tm = (tl + tr) / 2;
    if (idx <= tm)
        return t[v] + query(v * 2, tl, tm, idx);
    else
        return t[v] + query(v * 2 + 1, tm + 1, tr, idx);
}

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

    cin >> N >> Q >> S >> T;
    for (int i = 0; i <= N; i++)
        cin >> A[i];
    build(1, 0, N);

    ll res = 0;
    for (int i = 0; i < N; i++)
        res += max(0LL, A[i] - A[i + 1]) * T - max(0LL, A[i + 1] - A[i]) * S;

    int L, R;
    ll x;
    while (Q--)
    {
        cin >> L >> R >> x;

        A[L - 1] = query(1, 0, N, L - 1);
        A[R + 1] = query(1, 0, N, R + 1);
        A[L] = query(1, 0, N, L);
        A[R] = query(1, 0, N, R);

        res -= T * max(0LL, A[L - 1] - A[L]) - S * max(0LL, A[L] - A[L - 1]);
        if (R != N)
            res -= T * max(0LL, A[R] - A[R + 1]) - S * max(0LL, A[R + 1] - A[R]);

        add(1, 0, N, L, R, x);

        A[L] = query(1, 0, N, L);
        A[R] = query(1, 0, N, R);

        res += T * max(0LL, A[L - 1] - A[L]) - S * max(0LL, A[L] - A[L - 1]);
        if (R != N)
            res += T * max(0LL, A[R] - A[R + 1]) - S * max(0LL, A[R + 1] - A[R]);

        cout << res << '\n';
    }
    
    return 0;
}

• Subtask 1: 30 / 30
• Subtask 2: 10 / 10
• Subtask 3: 60 / 60