Submission #1160513

#	Time	Username	Problem	Language	Result	Execution time	Memory
1160513		steveonalex	Minerals (JOI19_minerals)	C++20	100 / 100	26 ms	4112 KiB

minerals

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define MASK(i) (1ULL << (i))
#define GETBIT(mask, i) (((mask) >> (i)) & 1)
#define ALL(v) (v).begin(), (v).end()

ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll gcd(ll a, ll b){return __gcd(a, b);}
ll lcm(ll a, ll b){return a / gcd(a, b) * b;}

ll LASTBIT(ll mask){return (mask) & (-mask);}
int pop_cnt(ull mask){return __builtin_popcountll(mask);}
int ctz(ull mask){return __builtin_ctzll(mask);}
int logOf(ull mask){return 63 - __builtin_clzll(mask);}

//mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
mt19937_64 rng(1);
ll rngesus(ll l, ll r){return l + (ull) rng() % (r - l + 1);}
double rngesus_d(double l, double r){
    double wow = (double) ((ull) rng()) / ((ull)(0-1));
    return wow * (r - l) + l;
}

template <class T1, class T2>
    bool maximize(T1 &a, T2 b){
        if (a < b) {a = b; return true;}
        return false;
    }

template <class T1, class T2>
    bool minimize(T1 &a, T2 b){
        if (a > b) {a = b; return true;}
        return false;
    }

template <class T>
    void printArr(T container, string separator = " ", string finish = "\n", ostream &out = cout){
        for(auto item: container) out << item << separator;
        out << finish;
    }

template <class T>
    void remove_dup(vector<T> &a){
        sort(ALL(a));
        a.resize(unique(ALL(a)) - a.begin());
    }

/*
Editorial for future me:
First, split the set into two part, then do your normal divide and conquer to get 40%
- Optimization 1: If the element is present in set 1, erasing or inserting will be the same in set 2
- Optimization 2: If one set is already filled up, skip
- Optimization 3: Shuffle to avoid hack
- Optimization 4: Also skip if you know it appear before mid
- Optimization 5: For large range, set ratio to a number different from 2
*/

#include "minerals.h"

pair<int, int> P;
vector<int> rape_victim;

void dnc(vector<int> &X, vector<int> Y, vector<int> &ans, int l, int r){
    if (l == r){
        ans[Y[0]] = X[l];
        return;
    }

    int mid;
    int cnt = 0;
    int R = 2;
    if (r - l + 1 > 100) R = 3;
    if (P == make_pair(l, r)){
        mid = r - (r - l + 1) / R;
        for(int i = mid + 1; i <= r; ++i)
            cnt = Query(X[i]);
    }
    else{
        mid = l + (r - l + 1) / R - 1;
        for(int i = l; i <= mid; ++i)
            cnt = Query(X[i]);
    }
    P = {l, mid};

    vector<int> left, right;
    for(int i: Y) {
        if (rape_victim[i] <= mid) {
            left.push_back(i);
            continue;
        }
        if (left.size() == mid - l + 1) {
            right.push_back(i);
            continue;
        }
        if (right.size() == r - mid){
            left.push_back(i);
            continue;
        }
        int _cnt = Query(i);
        if (_cnt == cnt) left.push_back(i);
        else right.push_back(i);

        cnt = _cnt;
    }

    dnc(X, left, ans, l, mid);
    dnc(X, right, ans, mid+1, r);
}

void Solve(int n) {
    vector<int> perm(n*2+1);
    for(int i = 1; i <= n*2; ++i) perm[i] = i;

    rape_victim.resize(n * 2 + 1);

    shuffle(1 + ALL(perm), rng);
    vector<int> X, Y;
    for(int it = 1; it <= n * 2; ++it){
        int i = perm[it];
        X.push_back(i);
        if (Query(i) != X.size()){
            X.pop_back();
            Y.push_back(i);
            rape_victim[i] = X.size();
        }
    }


    P = {0, n-1};
    vector<int> ans(n*2+1, -1);
    dnc(X, Y, ans, 0, n-1);
    for(int i = 1; i <= n*2; ++i) if (ans[i] != -1) Answer(i, ans[i]);
}

• Subtask 1: 6 / 6
• Subtask 2: 25 / 25
• Subtask 3: 9 / 9
• Subtask 4: 30 / 30
• Subtask 5: 5 / 5
• Subtask 6: 5 / 5
• Subtask 7: 5 / 5
• Subtask 8: 5 / 5
• Subtask 9: 10 / 10