Submission #581149

#	Time	Username	Problem	Language	Result	Execution time	Memory
581149		LucaIlie	Type Printer (IOI08_printer)	C++17	10 / 100	39 ms	2468 KiB

printer

#include <bits/stdc++.h>

#define MAX_N 25000
#define MAX_LEN 20

using namespace std;

int lcp[MAX_N], len[MAX_N];
string s[MAX_N];
vector <char> sol;

int main() {
    int n, minLen, first, last, i, j, l;
    bool cresc;

    cin >> n;
    for ( i = 0; i < n; i++ ) {
        cin >> s[i];
        len[i] = (int)s[i].size();
    }

    sort( s, s + n );

    for ( i = 0; i < n; i++ ) {
        j = (i + 1) % n;
        l = 0;
        while ( l < len[i] && l < len[j] && s[i][l] == s[j][l] )
            l++;
        lcp[i] = l;
    }

    minLen = MAX_LEN + 1;
    last = -1;
    cresc = true;
    for ( i = 0; i < n; i++ ) {
        if ( 2 * lcp[i] - len[i] < minLen ) {
            minLen = 2 * lcp[i] - len[i];
            last = i;
            cresc = true;
        }
        if ( 2 * lcp[(i - 1 + n) % n] - len[i] < minLen ) {
            minLen = 2 * lcp[(i - 1 + n) % n] - len[i];
            last = i;
            cresc = false;
        }
    }

    if ( cresc ) {
        first = (last + 1) % n;
        l = 0;
        for ( int k = 0; k < n; k++ ) {
            i = (first + k) % n;
            j = (i - 1 + n) % n;
            while ( l > lcp[j] ) {
                sol.push_back( '-' );
                l--;
            }
            while ( l < len[i] ) {
                sol.push_back( s[i][l] );
                l++;
            }
            sol.push_back( 'P' );
        }
    } else {
        first = (last - 1 + n) % n;
        l = 0;
        for ( int k = 0; k < n; k++ ) {
            i = (first - k + n) % n;
            j = (i - 1 + n) % n;
            while ( l > lcp[j] ) {
                sol.push_back( '-' );
                l--;
            }
            while ( l < len[i] ) {
                sol.push_back( s[i][l] );
                l++;
            }
            sol.push_back( 'P' );
        }
    }

    cout << sol.size() << "\n";
    for ( char o: sol )
        cout << o << "\n";

    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 0 / 10
• Subtask 3: 0 / 10
• Subtask 4: 0 / 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