제출 #901584

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
901584		abcvuitunggio	Shortcut (IOI16_shortcut)	C++17	0 / 100	1 ms	6744 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],idx[mxn],order[mxn];
T st[mxn*4];
vector <long long> v;
T operator +(T a, T b){
    return {max(a.first,b.first),min(a.second,b.second)};
}
void update(int node, int l, int r, int i){
    if (l==r){
        st[node]={s[l]+d[l],s[l]-d[l]};
        return;
    }
    int mid=(l+r)>>1;
    if (mid<i)
        update(node<<1|1,mid+1,r,i);
    else
        update(node<<1,l,mid,i);
    st[node]=st[node<<1]+st[node<<1|1];
}
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(node<<1,l,mid,u,v)+get(node<<1|1,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 make_pair(s[i]-d[i],-i)>make_pair(s[j]-d[j],-j);});
    vector <int> vd;
    for (int i=n-1;i>=0;i--)
        if (vd.empty()||vd.back()<order[i])
            vd.push_back(order[i]);
    reverse(vd.begin(),vd.end());
    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=0;
        for (int i=1;i<=n*4;i++)
            st[i]={-INF,INF};
        for (int i:vd){
            while (pos<n&&s[idx[pos]]+d[idx[pos]]>s[i]-d[i]+mid){
                update(1,0,n-1,idx[pos]);
                pos++;
            }
            if (!pos)
                continue;
            T t=get(1,0,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: 0 / 9
• Subtask 2: 0 / 14
• Subtask 3: 0 / 8
• Subtask 4: 0 / 7
• Subtask 5: 0 / 33
• Subtask 6: 0 / 22
• Subtask 7: 0 / 4
• Subtask 8: 0 / 3