Submission #433050

#	Time	Username	Problem	Language	Result	Execution time	Memory
433050		MarcoMeijer	Shortcut (IOI16_shortcut)	C++14	93 / 100	2082 ms	22068 KiB

shortcut

#include "shortcut.h"
#include <bits/stdc++.h>
using namespace std;

// macros
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> lll;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<lll> vlll;
#define REP(a,b,c) for(int a=int(b); a<int(c); a++)
#define RE(a,c) REP(a,0,c)
#define RE1(a,c) REP(a,1,c+1)
#define REI(a,b,c) REP(a,b,c+1)
#define REV(a,b,c) for(int a=int(c-1); a>=int(b); a--)
#define FOR(a,b) for(auto& a : b)
#define all(a) a.begin(), a.end()
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
#define sz size()
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll INF = 1e18;
const int N = 3e5+10;

ll n, c;
vll l, d;

struct MaxSeg {
    ll seg[N];
    void build(ll v) {
        RE(i,N) seg[i] = v;
    }
    void set(int i, ll v) {
        for(i++; i<N; i+=i&-i)
            seg[i] = max(seg[i], v);
    }
    ll get(int i) {
        ll res = -INF;
        for(i++; i>0; i-=i&-i)
            res = max(res, seg[i]);
        return res;
    }
} max1, max2;

vll difValues;
vi rval;

void preCalc() {
    difValues.assign(n,0);
    RE(i,n) difValues[i] = i;
    sort(all(difValues),[&](int i, int j) {
        return d[i]-l[i] > d[j]-l[j];
    });

    rval.assign(n,0);
    RE(i,n) rval[difValues[i]] = i;
}

bool possible(ll dist) {
    ll minSum=0, maxSum=INF, minDif=0, maxDif=INF;

    max1.build(-INF);
    max2.build(-INF);

    RE(i,n) {
        ll minValue = dist - d[i] - l[i];
        int lb=0, ub=n;
        while(lb != ub) {
            int mid=(lb+ub+1)/2;
            int j = difValues[mid-1];
            if(d[j]-l[j] > minValue) lb=mid;
            else ub=mid-1;
        }

        ll res1 = max1.get(lb-1);
        ll res2 = max2.get(lb-1);

        minDif = max(minDif, l[i] + d[i] - dist + c + res1);
        maxDif = min(maxDif, l[i] - d[i] + dist - c - res2);
        minSum = max(minSum, l[i] + d[i] - dist + c + res2);
        maxSum = min(maxSum, l[i] - d[i] + dist - c - res1);

        max1.set(rval[i],  d[i]-l[i]);
        max2.set(rval[i],  d[i]+l[i]);
    }

    REP(i,1,n) {
        ll curDif, curSum;
        int j1 = 0, j2 = 0;

        int lb=0, ub=i-1;
        while(lb != ub) {
            int mid=(lb+ub)/2;
            curDif = l[i]-l[mid];
            if(curDif > maxDif) lb=mid+1;
            else ub=mid;
        }
        j1 = lb;

        lb=0, ub=i-1;
        while(lb != ub) {
            int mid=(lb+ub)/2;
            curSum = l[i]+l[mid];
            if(curSum < minSum) lb=mid+1;
            else ub=mid;
        }
        j2 = lb;

        curDif = l[i]-l[j1];
        curSum = l[i]+l[j1];
        if(minDif <= curDif && curDif <= maxDif && minSum <= curSum && curSum <= maxSum)
            return true;

        curDif = l[i]-l[j2];
        curSum = l[i]+l[j2];
        if(minDif <= curDif && curDif <= maxDif && minSum <= curSum && curSum <= maxSum)
            return true;
    }
    return false;
}

ll find_shortcut(int n_, vi l_, vi d_, int c_) {
    n = n_; c=c_;
    l.pb(0);
    FOR(i,l_)
        l.pb(i);
    FOR(i,d_) d.pb(i);
    REP(i,1,n) l[i]+=l[i-1];

    preCalc();

    ll lb=0, ub=INF;
    while(lb != ub) {
        ll mid=(lb+ub)/2;
        if(possible(mid)) ub=mid;
        else lb=mid+1;
    }

    return lb;
}

• 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: 0 / 4
• Subtask 8: 0 / 3