Submission #901491

#	Time	Username	Problem	Language	Result	Execution time	Memory
901491		abcvuitunggio	Shortcut (IOI16_shortcut)	C++17	97 / 100	657 ms	524288 KiB

shortcut

#include "shortcut.h"
#include <bits/stdc++.h>
#define T pair <long long, long long>
using namespace std;
const long long INF=1LL<<51;
const int mxn=1000001;
long long s[mxn];
int d[mxn],id,idx[mxn],root[mxn],le[mxn*20],ri[mxn*20],order[mxn];
T st[mxn*20];
vector <long long> v;
T operator +(T a, T b){
    return {max(a.first,b.first),min(a.second,b.second)};
}
int update(int node, int l, int r, int i){
    id++;
    if (l==r){
        st[id]={s[l]+d[l],s[l]-d[l]};
        return id;
    }
    int cur=id,mid=(l+r)>>1;
    st[cur]=st[node];
    le[cur]=le[node];
    ri[cur]=ri[node];
    if (mid<i)
        ri[cur]=update(ri[node],mid+1,r,i);
    else
        le[cur]=update(le[node],l,mid,i);
    st[cur]=st[le[cur]]+st[ri[cur]];
    return cur;
}
T get(int node, int l, int r, int u, int v){
    if (l>v||r<u||l>r||u>v||!node)
        return {-INF,INF};
    if (l>=u&&r<=v)
        return st[node];
    int mid=(l+r)>>1;
    return get(le[node],l,mid,u,v)+get(ri[node],mid+1,r,u,v);
}
long long find_shortcut(int n, vector <int> l, vector <int> D, int c){
    for (int i=0;i<n;i++)
        d[i]=D[i];
    for (int i=1;i<n;i++){
        s[i]=s[i-1]+l[i-1];
        v.push_back(s[i]);
    }
    s[n]=INF;
    sort(v.begin(),v.end());
    iota(idx,idx+n,0);
    iota(order,order+n,0);
    sort(idx,idx+n,[](int i, int j){return s[i]+d[i]>s[j]+d[j];});
    sort(order,order+n,[](int i, int j){return s[i]-d[i]<s[j]-d[j];});
    st[0]={-INF,INF};
    for (int i=0;i<n;i++)
        root[i]=update((i?root[i-1]:0),0,n-1,idx[i]);
    long long lo=0,hi=INF,kq=-1;
    while (lo<=hi){
        long long mid=(lo+hi)>>1,ch=0,mnx=-INF,mxx=INF,mny=-INF,mxy=INF;
        int pos=n-1;
        for (int j=0;j<n;j++){
            int i=order[j];
            while (pos>=0&&s[idx[pos]]+d[idx[pos]]<=s[i]-d[i]+mid)
                pos--;
            if (pos<0)
                break;
            T t=get(root[pos],0,n-1,i+1,n-1);
            mnx=max(mnx,s[i]+d[i]+c-mid+t.first);
            mxx=min(mxx,s[i]-d[i]-c+mid+t.second);
            mny=max(mny,s[i]+d[i]+c-mid-t.second);
            mxy=min(mxy,s[i]-d[i]-c+mid-t.first);
        }
        int tmp=0,j=n;
        for (int i=0;i<n;i++){
            long long l=max(mnx-s[i],s[i]-mxy),r=min(mxx-s[i],s[i]-mny);
            if (mnx-s[i]<s[i]-mxy&&!tmp){
                j=0;
                tmp=1;
            }
            if (tmp)
                while (j<n&&s[j]<l)
                    j++;
            else{
                while (j>=0&&s[j]>=l)
                    j--;
                j++;
            }
            if (j<n&&s[j]<=r){
                ch=1;
                break;
            }
        }
        if (ch){
            kq=mid;
            hi=mid-1;
        }
        else
            lo=mid+1;
    }
    return kq;
}

• Subtask 1: 9 / 9
• Subtask 2: 14 / 14
• Subtask 3: 8 / 8
• Subtask 4: 7 / 7
• Subtask 5: 33 / 33
• Subtask 6: 22 / 22
• Subtask 7: 4 / 4
• Subtask 8: 0 / 3