Submission #660057

#	Time	Username	Problem	Language	Result	Execution time	Memory
660057		Mahdi	The short shank; Redemption (BOI21_prison)	C++17	0 / 100	1 ms	340 KiB

prison

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
#define F first
#define S second
typedef long long ll;
typedef pair<int, int> pii;
const int N=2e6+1e5;
int n, d, t, a[N], mx[N], nx[N], be[N], h[N];
set<int>zr;

struct fenwick{
    int bit[N];

    void add(int i, int val=1){
        for(;i<n;i|=(i+1))
            bit[i]+=val;
    }

    int sum(int r){
        int res=0;
        for(;r>=0;r=(r&(r+1))-1)
            res+=bit[r];
        return res;
    }

    int sum(int l, int r){
        if(r>l)
            return sum(r-1)-sum(l-1);
        return 0;
    }
} A;

inline void del(int l, int r){
    auto it=zr.lower_bound(l);
    while(it!=zr.end() && *it<r){
        A.add(*it, -1);
        zr.erase(it);
        it=zr.lower_bound(l);
    }
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin>>n>>d>>t;
    for(int i=0;i<n;++i)
        cin>>a[i];
    int ans=0;
    for(int i=0;i<n;++i){
        mx[i]=t-a[i]+i+1;
        if(a[i]<=t)
            ++ans;
        else{
            zr.insert(i);
            A.add(i);
        }
    }
    for(int i=0;i<n-1;++i){
        if(mx[i]>i+1 && mx[i+1]<=i+1)
            h[i]=1;
    }
    iota(nx, nx+n, 1);
    iota(be, be+n, -1);
    set<pii>st;
    for(int i=0;i<n-1;++i){
        st.insert({h[i], i});
    }
    for(int i=n-1;i>d;--i){
        pii z=*st.begin();
        st.erase(st.begin());
        ans+=z.F;
        int r=nx[z.S];
        int l=be[z.S];
        del(z.S+1, min(mx[z.S], r+1));
        be[r]=l;
        if(l>=0)
            nx[l]=r;
        if(r!=n-1){
            mx[r]=max(mx[r], mx[z.S]);
            st.erase({h[r], r});
            h[r]=A.sum(r+1, min(nx[r]+1, mx[r]));
            st.insert({h[r], r});
        }
        if(l!=-1){
            st.erase({h[l], l});
            h[r]=A.sum(l+1, min(r+1, mx[l]));
            st.insert({h[l], l});
        }
    }
    cout<<ans<<'\n';
}

• Subtask 1: 0 / 15
• Subtask 2: 0 / 10
• Subtask 3: 0 / 20
• Subtask 4: 0 / 10
• Subtask 5: 0 / 25
• Subtask 6: 0 / 20