Submission #881636

#	Time	Username	Problem	Language	Result	Execution time	Memory
881636		shleep	Type Printer (IOI08_printer)	C++17	30 / 100	71 ms	72904 KiB

printer

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

// Trie
vector<vector<int>> nex(500001, vector<int>(26));
vector<int> endpoint(500001); // marks if a word is done
vector<int> depth(500001, -1);
int num_nodes = 0;

void trie_insert(string s) {
    int cur = 0;
    for (char c : s) {
        if (!nex[cur][c-'a'])
            nex[cur][c-'a'] = ++num_nodes;
        cur = nex[cur][c-'a'];
    }
    endpoint[cur]=1;
}

// Find max depth of all nodes in subtree
int depth_dfs(int cur) {
    if (endpoint[cur]) {
        depth[cur]=1;
        return 1;
    }
    int deepest = 0;
    for (int i=0;i<26;i++) {
        if (!nex[cur][i]) continue;
        deepest = max(deepest, depth_dfs(nex[cur][i]));
    }
    depth[cur] = deepest+1;
    return deepest+1;
}

int main() {
    int N; cin>>N;
    vector<string> S(N);
    for (int i=0;i<N;i++)
        cin >> S[i];
    
    for (string s : S)
        trie_insert(s);
    
    // Calculate depths of every node
    depth_dfs(0);
    
    // Perform backtracking dfs, while prioritizing shorter depths
    // (this ensures we end on the longest word, which minimizes moves)
    vector<char> ans;
    auto dfs = [&](auto &&self, int cur) -> void {
        // end of word (but dont terminate early, e.g. "cat,cats")
        if (endpoint[cur]) {
            ans.push_back('P');
        }
        // Visit child nodes in order of their depths
        vector<array<int,3>> candidates;
        for (int i=0;i<26;i++) {
            if (!nex[cur][i]) continue;
            int nxt = nex[cur][i];
            candidates.push_back({depth[nxt], nxt, i});
        }
        sort(candidates.begin(),candidates.end());

        for (auto [_,nxt,chr] : candidates) {
            ans.push_back((char)(chr+'a'));
            self(self, nxt);
            ans.push_back('-');
        }
    };
    dfs(dfs, 0);

    // Get rid of final '-'s (we don't need to clean up)
    while (ans.back() == '-')
        ans.pop_back();

    // Print results
    cout << ans.size() << '\n';
    for (char c : ans)
        cout << c << '\n';
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 0 / 10
• Subtask 4: 10 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10