제출 #1074826

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1074826		dosts	Global Warming (CEOI18_glo)	C++17	100 / 100	419 ms	110784 KiB

glo

//Dost SEFEROĞLU
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pii pair<int,int>
#define ff first
#define ss second
#define sp << " " <<    
#define all(cont) cont.begin(),cont.end()
#define vi vector<int>
const int MOD = 1e9+7,inf = 2e12;
const int N = 5001;
vi vv;

int idx(int x) {
    return upper_bound(all(vv),x)-vv.begin();
}

struct FT{
    int n;
    vi bit;

    FT(int nn) {
        n = nn;
        bit.assign(nn+1,0);
    }

    void put(int p,int v) {
        for (int i=p;i<=n;i+=i&-i) bit[i]=max(bit[i],v);
    } 

    int get(int p) {
        int ans = 0;
        for (int i=p;i>0;i-=i&-i) ans=max(ans,bit[i]);
        return ans;
    }
    void reset() {
        bit.assign(n+1,0);
    }
};

struct ST {
    vector<multiset<int,greater<int>>> t;
    vector<int> ans;
    ST(int nn) {
        t.resize(4*nn+1);
        ans.resize(4*nn+1);
    }

    void insert(int node,int l,int r,int p,int v) {
        if (l == r) {
            t[node].insert(v);
            ans[node] = *t[node].begin();
            return;
        }
        int m = (l+r) >> 1;
        if (p<=m) insert(2*node,l,m,p,v);
        else insert(2*node+1,m+1,r,p,v);
        ans[node] = max(ans[2*node],ans[2*node+1]);
    }
    void erase(int node,int l,int r,int p,int v) {
        if (l == r) {
            t[node].erase(t[node].find(v));
            if (t[node].empty()) ans[node] = 0;
            else ans[node] = *t[node].begin();
            return;
        }
        int m = (l+r) >> 1;
        if (p<=m) erase(2*node,l,m,p,v);
        else erase(2*node+1,m+1,r,p,v);   
        ans[node] = max(ans[2*node],ans[2*node+1]);
    }
    int query(int node,int l,int r,int L,int R) {
        if (l > R || r < L) return 0;
        if (l >= L && r<=R) return ans[node];
        int m = (l+r) >> 1;
        return max(query(2*node,l,m,L,R),query(2*node+1,m+1,r,L,R));
    }
};
void solve() { 
    int n,x;
    cin >> n >> x;
    //corollary: always decrease some prefix
    //corollary: if you decrease 1..i you MUST use i.
    vi a(n+1);
    for (int i=1;i<=n;i++) cin >> a[i];
    for (int i=1;i<=n;i++) {
        vv.push_back(a[i]);
        vv.push_back(a[i]-x);
    }
    sort(all(vv));
    vv.erase(unique(all(vv)),vv.end());
    int sz = vv.size();
    FT ft(sz);
    vi p(n+1),s(n+1);
    for (int i=1;i<=n;i++) {
        p[i] = ft.get(idx(a[i])-1)+1;
        ft.put(idx(a[i]),p[i]);
    }
    ft.reset();
    for (int i=n;i>=1;i--) {
        s[i] = ft.get(sz-idx(a[i]))+1;
        ft.put(sz-idx(a[i])+1,s[i]);
    }
    ft.reset();
    int ans = 0;
    ST st(sz);
    for (int i=1;i<=n;i++) st.insert(1,1,sz,idx(a[i]),s[i]);
    for (int i=1;i<=n;i++) {
        st.erase(1,1,sz,idx(a[i]),s[i]);
        ans = max(ans,p[i]+st.query(1,1,sz,idx(a[i]-x)+1,sz));
    }
    cout << ans << endl;
}
 
signed main() { 
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    #ifdef Dodi
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int t = 1;
    //cin >> t; 
    while (t --> 0) solve();
}

• Subtask 1: 5 / 5
• Subtask 2: 10 / 10
• Subtask 3: 13 / 13
• Subtask 4: 10 / 10
• Subtask 5: 20 / 20
• Subtask 6: 17 / 17
• Subtask 7: 25 / 25