제출 #444391

#제출 시각아이디문제언어결과실행 시간메모리
444391JvThunder사탕 분배 (IOI21_candies)C++17
100 / 100
1345 ms48292 KiB
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
 
const int MAXQ = 1000005;
const ll INF = (ll)(1e18);
 
struct SegmentTree 
{
    int n;
    ll vmin[MAXQ] = {0};
    ll vmax[MAXQ] = {0};
    ll lazy[MAXQ] = {0};
 
    void init(int _n) { n = _n; }
 
    void lazy_update(int node, int from, int to)
    {
        vmin[node] += lazy[node]; vmax[node] += lazy[node];
        if (from < to) 
        {
            lazy[2*node] += lazy[node];
            lazy[2*node+1] += lazy[node];
        }
        lazy[node] = 0;
    }
 
    void update(int node, int from, int to, int L, int R, int add)
    {
        lazy_update(node, from, to);
        if (from > R || to < L) return;
        if (L <= from && to <= R) {
            lazy[node] += add;
            lazy_update(node, from, to);
            return;
        }
        int mid = (from + to) / 2;
        update(node * 2, from, mid, L, R, add);
        update(node * 2 + 1, mid + 1, to, L, R, add);
        vmin[node] = min(vmin[node * 2], vmin[node * 2 + 1]);
        vmax[node] = max(vmax[node * 2], vmax[node * 2 + 1]);
    }
 
    ll get(int node, int from, int to, int lowerbound) 
    {
        lazy_update(node, from, to);
        if (from == to) return vmin[node];
        int mid = (from + to) / 2;
        if (vmax[node * 2 + 1] + lazy[node * 2 + 1] >= lowerbound)
            return get(node * 2 + 1, mid + 1, to, lowerbound);
        return min( vmin[node * 2 + 1] + lazy[node * 2 + 1], get(node * 2, from, mid, lowerbound) );
    }
 
    void add_range(int L, int R, int add) 
    {
        update(1, 0, n - 1, L, R, add);
    }
 
    ll min_suffix(int lowerbound) 
    {
        if (vmax[1] < lowerbound) return -INF;
        return min(0LL, get(1, 0, n - 1, lowerbound));
    }
} T;
 
vector<int> distribute_candies(vector<int> C, vector<int> L, vector<int> R, vector<int> V) 
{
    int n = C.size(); int q = L.size();
    vector<int> A(n);
    
    vector<vector<int>> begin_at(n,vector<int>()), end_at(n,vector<int>());
    for(int i=0;i<q;i++) 
    {
        begin_at[L[i]].push_back(i);
        end_at[R[i]].push_back(i);
    }
 
    T.init(q+1);
    vector<int> final_A(n);
    for(int i=0;i<n;i++)
    {
        if (i>0) 
        {
            T.add_range(0,0,-A[i-1]);
            for(int j:end_at[i-1]) T.add_range(0,1+j,-V[j]);
        }
        T.add_range(0,0,A[i]);
        for(int j : begin_at[i]) T.add_range(0,1+j,V[j]);
 
        int low = 1, high = C[i] + 1;
        while (low <= high) 
        {
            int mid = (low + high) / 2;
            ll smin = T.min_suffix(mid);
            if (-smin + mid > C[i]) high = mid - 1;
            else low = mid + 1;
        }
        final_A[i] = high;
    }
 
    return final_A;
}
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