제출 #817773

#제출 시각아이디문제언어결과실행 시간메모리
817773t6twotwoMeasures (CEOI22_measures)C++17
76 / 100
309 ms30556 KiB
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define FFOR(i,a,b,c) for(int i=(a);(c)>0?i<=(b):i>=(b);i+=(c))
#define FOOR(i,a,b) FFOR(i,a,(b)-1,1)
#define ROOF(i,a,b) FFOR(i,(b)-1,a,-1)
#define FOR(i,n) FOOR(i,0,n)
#define ROF(i,n) ROOF(i,0,n)
#define FORR(x,v) for(auto &x:v)
#define vc vector
#define sz(v) (int)((v).size())
#define bg begin()
#define en end()
#define em empty()
#define lb lower_bound
#define ub upper_bound
#define pb push_back
#define ppb pop_back()
#define eb emplace_back
#define bk back()
#define fr front()
#define pp pop()
#define all(v) (v).begin(),(v).end()
#define lla(v) (v).rbegin(),(v).rend()
#define f first
#define s second
template <class T>
struct segtree {
    int n;
    vector<T> s;
    segtree(int n) : n(n), s(2 * n) {
    }
    segtree(vector<T> &a) : segtree(a.size()) {
        for (int i = 2 * n - 1; i; i--) {
            s[i] = i < n ? s[i * 2] + s[i * 2 + 1] : a[i - n];
        }
    }
    void update(int i, const T &v) {
        for (s[i += n] = v; i /= 2;) {
            s[i] = s[i * 2] + s[i * 2 + 1];
        }
    }
    T query(int l, int r) {
        T u, v;
        for (l += n, r += n; l < r; l /= 2, r /= 2) {
            if (l % 2 == 1) {
                u = u + s[l++];
            }
            if (r % 2 == 1) {
                v = s[--r] + v;
            }
        }
        return u + v;
    }
};
struct node {
    ll sum, pre, suf, ans;
    node() {
        sum = pre = suf = ans = 0;
    }
    node(int v) {
        sum = v;
        pre = suf = ans = max(0, v);
    }
};
node operator+(const node &a, const node &b) {
    node c;
    c.sum = a.sum + b.sum;
    c.pre = max(a.pre, a.sum + b.pre);
    c.suf = max(b.suf, b.sum + a.suf);
    c.ans = max({a.suf + b.pre, a.ans, b.ans});
    return c;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N,M,D;
    cin>>N>>M>>D;
    if(M<=10&&0){
        vc<int> a(N);
        FORR(x,a)cin>>x;
        FOR(i,M){
            int b;
            cin>>b;
            a.pb(b);
            sort(all(a));
            ll s=0,ans=0;
            FOOR(j,1,sz(a)){
                ll k=a[j]-a[j-1]-D;
                s=min(k,s+k);
                ans=min(ans,s);
            }
            ans=-ans;
            if(ans%2==1){
                cout<<ans/2<<"."<<5<<" ";
            }else{
                cout<<ans/2<<" ";
            }
        }
        return 0;
    }
    vc<int> b(M);
    vc<pair<int,int>> p(M);
    FOR(i,M){
        cin>>b[i];
        p[i]={b[i],i};
    }
    sort(all(p));
    vc<int> pos(M);
    FOR(i,M) pos[p[i].s]=i;
    segtree<node> st(M);
    set<int> s;
    FOR(i,M){
        auto it=s.ub(pos[i]);
        if(it!=s.bg){
            st.update(pos[i],-(b[i]-p[*prev(it)].f-D));
        }
        if(it!=s.en){
            st.update(*it,-(p[*it].f-b[i]-D));
        }
        s.insert(pos[i]);
        ll ans=st.query(0,M).ans;
        if(ans%2==1){
            cout<<ans/2<<"."<<5<<" ";
        }else{
            cout<<ans/2<<" ";
        }
    }
    return 6/22;
}
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