Submission #1212655

#	Time	Username	Problem	Language	Result	Execution time	Memory
1212655		onbert	Hieroglyphs (IOI24_hieroglyphs)	C++20	16 / 100	44 ms	21068 KiB

hieroglyphs

#include "hieroglyphs.h"
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5, maxval = 2e5 + 5;
int n, m;
vector<int> a, b;
int afreq[maxval], bfreq[maxval];
vector<int> apos[maxval], bpos[maxval];
vector<pair<int,int>> acrit, bcrit;
bool aiscrit[maxn], biscrit[maxn];
int asz, bsz;
int l[maxn], r[maxn];
vector<int> vec[maxn];

vector<int> ucs(vector<int> A, vector<int> B) {
    for (int i:A) afreq[i]++;
    for (int i:B) bfreq[i]++;
    a.push_back(2e5 + 1), b.push_back(2e5 + 1);
    for (int i:A) if (bfreq[i]) a.push_back(i);
    for (int i:B) if (afreq[i]) b.push_back(i);
    a.push_back(2e5 + 2), b.push_back(2e5 + 2);

    n = a.size(), m = b.size();
    for (int i=0;i<n;i++) if (afreq[a[i]] < bfreq[a[i]]) {
        acrit.push_back({a[i], i});
        aiscrit[i] = true;
    }
    for (int i=0;i<m;i++) if (bfreq[b[i]] <= afreq[b[i]]) {
        bcrit.push_back({b[i], i});
        biscrit[i] = true;
    }
    asz = acrit.size(), bsz = bcrit.size();

    for (int i=0;i<n;i++) apos[a[i]].push_back(i);
    for (int i=0;i<m;i++) bpos[b[i]].push_back(i);

    int pt = bsz;
    for (int i=n-1;i>=0;i--) {
        if (pt-1 >= 0 && bcrit[pt-1].first == a[i]) {
            pt--, r[pt] = i;
        }
    }

    vector<int> ans;
    int apt = -1, bpt = -1;
    for (int i=0;i<bsz;i++) {
        auto [val, pos] = bcrit[i];
        while (apt+1 < asz && acrit[apt+1].second < r[i] && bpt < pos) {
            bpt++;
            if (b[bpt] == acrit[apt+1].first) {
                vec[i-1].push_back({b[bpt]});
                ans.push_back(b[bpt]);
                apt++;
            }
        }
        if (i != 0 && i != bsz-1) ans.push_back(val);
        if (apt != -1) {
            int nxt = *upper_bound(apos[val].begin(), apos[val].end(), acrit[apt].second);
            if (apt+1 < asz && acrit[apt+1].second < nxt) return {-1};
        }
        bpt = pos;
    }
    return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 3
• Subtask 2: 0 / 15
• Subtask 3: 0 / 10
• Subtask 4: 16 / 16
• Subtask 5: 0 / 14
• Subtask 6: 0 / 42