Submission #660063

#	Time	Username	Problem	Language	Result	Execution time	Memory
660063		Mahdi	The short shank; Redemption (BOI21_prison)	C++17	80 / 100	2075 ms	149200 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;
            }
            int m=(lx+rx)/2;
            if(sm[x]==rx-lx)
                sm[2*x+1]=sm[2*x+2]=sm[x]>>1;
            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];
        }

        void one(int l, int r){
            if(r>l)
                one(0, 0, sz, l, r);
        }
        
        int sum(int x, int lx, int rx, const int &l, const 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]>>1;
            int m=(lx+rx)/2;
            return sum(2*x+1, lx, m, l, r)+sum(2*x+2, m, rx, l, r);
        }
        
        int sum(int l, int r){
            if(r>l)
                return sum(0, 0, sz, l, r);
            return 0;
        }
    } A;
     
    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];
        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