Submission #480892

#	Time	Username	Problem	Language	Result	Execution time	Memory
480892		blue	Board (CEOI13_board)	C++17	10 / 100	3 ms	1472 KiB

board

#include <iostream>
#include <vector>
#include <string>
#include <deque>
using namespace std;

const int MX = 100'000; //FIX

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    vector<int> mov[2];
    vector<int> depth(2, 0);

    for(int q = 0; q < 2; q++)
    {
        string A;
        cin >> A;

        mov[q] = vector<int>(1+MX, 0);

        for(char c: A)
        {
            if(c == '1') depth[q]++;
            else if(c == '2')
            {
                depth[q]++;
                mov[q][depth[q]]++;
            }
            else if(c == 'U') depth[q]--;
            else if(c == 'L') mov[q][depth[q]]--;
            else if(c == 'R') mov[q][depth[q]]++;
        }

        for(int d = MX; d >= 1; d--)
        {
            // cerr << d << ' ' << mov[q][d] << '\n';
            mov[q][d-1] += mov[q][d]/2;
            mov[q][d] -= 2*(mov[q][d]/2);
            if(mov[q][d] == -1)
            {
                mov[q][d-1]--;
                mov[q][d] += 2;
            }
        }

        // for(int h = 1; h <= depth[q]; h++)
        // {
        //     if(mov[q][h] == 0) cerr << "L";
        //     else cerr << "R";
        // }
        // cerr << "\n";
    }

    int X = 0, Y = 1;
    if(depth[X] > depth[Y]) swap(X, Y);

    //depth[X] <= depth[Y];

    int ans = 0;
    ans += depth[Y] - depth[X];
    depth[Y] = depth[X]; //Y := ancestor of Y at level of X.


    //X and Y are now at the same depth


    deque<int> M[2];
    for(int q = 0; q < 2; q++)
    {
        for(int v = 1; v <= depth[X]; v++)
        {
            M[q].push_back(mov[q][v]);
        }
    }

    //sequence of moves from root to both X and Y found.

    // for(int g = 0; g < int(M[0].size()); g++) cerr << M[0][g] << ' ' << M[1][g] << '\n';
    // cerr << "\n\n";

    while(!M[0].empty() && M[0][0] == M[1][0])
    {
        M[0].pop_front();
        M[1].pop_front();
    }

    // for(int g = 0; g < int(M[0].size()); g++) cerr << M[0][g] << ' ' << M[1][g] << '\n';
    // cerr << "\n\n";

    //sequence of moves up to LCA removed.

    if(M[0].empty())
    {
        cout << ans << '\n';
        return 0;
    }

    //if one is ancestor of other, return

    if(M[X][0] != 0) swap(X, Y);


    //now, X is on the left.

    // for(int g = 0; g < int(M[0].size()); g++) cerr << M[X][g] << ' ' << M[Y][g] << '\n';
    // cerr << "\n\n";

    M[X].pop_front();
    M[Y].pop_front();

    //sequence of moves from LCA-children to X and Y.

    // for(int g = 0; g < int(M[0].size()); g++) cerr << M[X][g] << ' ' << M[Y][g] << '\n';
    // cerr << "\n\n";

    // cerr << ans << '\n';

    int neq = -1;
    for(int i = 0; i < (int)M[X].size(); i++)
    {
        if(M[X][i] != 1 || M[Y][i] != 0)
        {
            neq = i;
            break;
        }
    }

    // cerr << neq << '\n';
    // for(int g = 0; g < int(M[0].size()); g++) cerr << M[X][g] << ' ' << M[Y][g] << '\n';
    // cerr << "\n\n";

    if(neq == -1)
    {
        cout << 1 + ans << '\n';
        return 0;
    }

    //if X and Y are already on subtree sides, return 1 + ans

    while(int(M[X].size()) - 5 > neq)
    {
        M[X].pop_back();
        M[Y].pop_back();
        ans += 2;
    }

    int newAns = 1'000'000;

    while(int(M[X].size()) - 1 >= neq)
    {
        int v1 = 0;
        int v2 = 1;
        for(int f = neq; f < int(M[X].size()); f++)
        {
            v1 = 2*v1 + M[X][f];
            v2 = 2*v2 + M[Y][f];
        }

        newAns = min(newAns, ans + (v2 - v1));

        ans += 2;
        M[X].pop_back();
        M[Y].pop_back();
    }

    cout << newAns << '\n';
}

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• Subtask 5: 10 / 10
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