Submission #724429

#	Time	Username	Problem	Language	Result	Execution time	Memory
724429		MohamedAliSaidane	Mars (APIO22_mars)	C++17	14 / 100	17 ms	2112 KiB

mars

#include<bits/stdc++.h>
#include "mars.h"
#include<ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;

typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int,int> pii;
typedef vector<pii> vpi;

#define ff first
#define ss second
#define pb push_back
#define all(x) (x).begin(),(x).end()



string process(vector<vector<string>> a, int i, int j, int k, int n)
{
    int nx[4] = {1, -1, 0, 0};
    int ny[4] = {0, 0, 1, -1};
    int bs[100];
    memset(bs, 0, sizeof(bs));
    if(k == 0)
    {
        for(int l = i; l < i + 3; l++)
        {
            for(int p = j; p < j+3; p++)
            {
                int id = (2 * n + 1) * l + p;
                bs[id] = (int)(a[l - i][p - j][0] - '0');
            }
        }
    }
    else
    {
        for(int l = i; l < i + 3; l++)
        {
            for(int p = j; p < j+3; p++)
            {
                for(int h = 0; h < 100; h++)
                    bs[h] = (bs[h]) |  (int)(a[l-i][p-j][h] - '0');
            }
        }
    }
    string ans = "";
    if(k == n -1)
    {
        for(int h = 0; h < 100; h++)
            ans += '0';

        int mp[2 * n + 1][2 * n + 1];
        memset(mp, 0, sizeof(mp));
        for(int h = 0; h < (2 * n + 1) * (2 * n + 1); h++)
        {
            mp[h/(2 * n + 1)][h % (2 * n + 1)] = bs[h];
        }
        int vis[2 * n + 1][2 * n + 1];
        memset(vis, 0, sizeof(vis));
        int cnt = 0;
        for(int l =0; l < 2 * n + 1;l ++)
        {
            for(int p = 0; p < 2 * n + 1; p++)
            {
                if(mp[l][p] & (1 - vis[l][p]))
                {
                    cnt++;
                    queue<pii> q;
                    q.push({l, p});
                    vis[l][p] = 1;
                    while(!q.empty())
                    {
                        pii u = q.front();
                        q.pop();
                        for(int h = 0; h < 4; h++)
                        {
                            int x = nx[h] + u.ff;
                            int y = ny[h] + u.ss;
                            if((x >= 0) && (x < 2 * n + 1) && (y >= 0) && (y < 2 * n + 1) &&
                               mp[x][y] && (1 - vis[x][y]))
                            {
                                vis[x][y] = 1;
                                q.push({x, y});
                            }
                        }
                    }
                }
            }
        }
        for(int h =0; h < 10; h++)
        {
            if((1 << h) & cnt)
                ans[h] = '1';
        }
    }
    else
    {
        for(int h = 0; h < 100; h++)
            ans += (char)('0' + (int)(bs[h]));
    }
    return ans;
}

• Subtask 1: 6 / 6
• Subtask 2: 8 / 8
• Subtask 3: 0 / 7
• Subtask 4: 0 / 8
• Subtask 5: 0 / 7
• Subtask 6: 0 / 8
• Subtask 7: 0 / 10
• Subtask 8: 0 / 24
• Subtask 9: 0 / 11
• Subtask 10: 0 / 11