제출 #1313180

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1313180		Tymond	상형문자열 (IOI24_hieroglyphs)	C++20	0 / 100	273 ms	23756 KiB

hieroglyphs

#include "hieroglyphs.h"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define fi first
#define se second
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int rand(int l,int r){return uniform_int_distribution<int>(l, r)(rng);}
inline ll rand(ll l,ll r){return uniform_int_distribution<ll>(l, r)(rng64);}
#ifdef DEBUG
auto&operator<<(auto&o,pair<auto,auto>p){return o<<"("<<p.first<<", "<<p.second<<")";}
auto operator<<(auto&o,auto x)->decltype(x.end(),o){o<<"{";int i=0;for(auto e:x)o<<","+!i++<<e;return o<<"}";}
#define debug(X...)cerr<<"["#X"]: ",[](auto...$){((cerr<<$<<"; "),...)<<endl;}(X)
#else
#define debug(...){}
#endif
vi ucs(vi A, vi B){
    int n = sz(A);
    int m = sz(B);
    vi ret = {-1};
    
    ret = {};
    //znajdź wynik
    set<int> ind = {};
    map<int, vi> posA;
    map<int, vi> posB;
    map<int, int> cntA;
    map<int, int> cntB;
    for(int i = n - 1; i >= 0; i--){
        posA[A[i]].pb(i);
        cntA[A[i]]++;
    }
    for(int i = m - 1; i >= 0; i--){
        posB[B[i]].pb(i);
        cntB[B[i]]++;
    }
    
    int mx = -1;
    for(int i = 0; i < n; i++){
        if(cntA[A[i]] > cntB[A[i]]){
            continue;
        }

        while(sz(posB[A[i]]) && posB[A[i]].back() <= mx){
            posB[A[i]].pop_back();
        }
        ind.insert(i);
        mx = posB[A[i]].back();
        posB[A[i]].pop_back();
    }

    mx = -1;
    for(int i = 0; i < m; i++){
        if(cntB[B[i]] >= cntA[B[i]]){
            continue;
        }

        while(sz(posA[B[i]]) && posA[B[i]].back() <= mx){
            posA[B[i]].pop_back();
        }
        ind.insert(posA[B[i]].back());
        mx = posA[B[i]].back();
        posA[B[i]].pop_back();
    }

    for(auto ele : ind){
        ret.pb(A[ele]);
    }

    return ret;
}

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