Submission #660034

#	Time	Username	Problem	Language	Result	Execution time	Memory
660034		Mahdi	The short shank; Redemption (BOI21_prison)	C++17	80 / 100	602 ms	37592 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];

struct seg{
    int sz, sm[2*N];
    
    void ok(){
        sz=1;
        while(sz<n)
            sz<<=1;
    }
    
    void one(int x, int lx, int rx, int l, int r){
        if(lx>=r || rx<=l)
            return;
        if(lx>=l && rx<=r){
            sm[x]=rx-lx;
            return;
        }
        if(sm[x]==rx-lx)
            sm[2*x+1]=sm[2*x+2]=sm[x]/2;
        int m=(lx+rx)/2;
        one(2*x+1, lx, m, l, r);
        one(2*x+2, m, rx, l, r);
        sm[x]=sm[2*x+1]+sm[2*x+2];
    }

    int sum(int x, int lx, int rx, int l, int r){
        if(lx>=r || rx<=l)
            return 0;
        if(lx>=l && rx<=r)
            return sm[x];
        if(sm[x]==rx-lx)
            sm[2*x+1]=sm[2*x+2]=sm[x]/2;
        int m=(lx+rx)/2;
        return sum(2*x+1, lx, m, l, r)+sum(2*x+2, m, rx, l, r);
    }
} A;

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin>>n>>d>>t;
    if(n>N/2)
        return 0;
    for(int i=0;i<n;++i)
        cin>>a[i];
    A.ok();
    int ans=0;
    for(int i=0;i<n;++i){
        mx[i]=t-a[i]+i+1;
        if(a[i]<=t){
            A.one(0, 0, A.sz, i, i+1);
            ++ans;
        }
    }
    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];
        A.one(0, 0, A.sz, 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]=max(0, min(nx[r], mx[r]-1)-r)-A.sum(0, 0, A.sz, r+1, min(nx[r]+1, mx[r]));
            st.insert({h[r], r});
        }
        if(l!=-1){
            st.erase({h[l], l});
            h[l]=max(0, min(r, mx[l]-1)-l)-A.sum(0, 0, A.sz, l+1, min(r+1, mx[l]));
            st.insert({h[l], l});
        }
    }
    cout<<ans<<'\n';
}

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