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CSES MONSTERS . Need help with code
By pratikshaw, history, 3 years ago, In English

Problem link : https://cses.fi/problemset/task/1194 Submission link: https://cses.fi/problemset/result/6625261/

I have done the problem in n*m which is the most optimised one, (1<=n,m<=10^3) . Even then I am getting TLE in one test case

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3 years ago, hide # |
 
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Start a BFS from E, and find shortest distance ($$$D_{N_{i}}$$$) to every node ($$$(N_{i}$$$) and if for any monster, there exists a ($$$D_{N_{i}}$$$) which is less than or equal to ($$$D_{N_{i}}$$$) from S to E, we can't reach the end point, otherwise we can.

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3 years ago, hide # |
 
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You can check this out. Hope it helps

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    3 years ago, hide # ^ |
     
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    I havent used maps , I have only used 2-D arrays

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      3 years ago, hide # ^ |
       
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      2-D arrays is much faster than maps, If u can implement my solution by replacing maps with 2-D arrays, then ur code's runtime would be lesser than mine and would get accepted at the very first try as I have to grind a lot to escape TLE.

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3 years ago, hide # |
 
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your submission link isn't working, it's blank.

I guess CSES submission link is private-

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    3 years ago, hide # ^ |
     
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    include "bits/stdc++.h"

    using namespace std;

    define ll long long

    int main() { ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); ll n, m; cin >> n >> m; vector<vector> arr(n, vector(m)); ll b = -1, c = -1; queue<pair<ll, ll>> q; vector<vector> visited(n, vector(m, false));

    for (ll i = 0; i < n; i++)
{
    for (ll j = 0; j < m; j++)
    {
        cin >> arr[i][j];
        if (arr[i][j] == 'A')
        {
            b = i;
            c = j;
        }
        else if (arr[i][j] == 'M')
        {
            q.push({i, j});
            visited[i][j] = true;
        }
    }
}

vector<vector<ll>> time(n, vector<ll>(m, INT_MAX));
ll steps = 0;
while (!q.empty())
{
    ll sz = q.size();
    while (sz--)
    {
        auto front = q.front();
        q.pop();
        ll x = front.first, y = front.second;
        time[x][y] = steps;

        if (x + 1 < n and !visited[x + 1][y] and arr[x + 1][y] == '.')
        {
            visited[x + 1][y] = true;
            q.push({x + 1, y});
        }
        if (x - 1 >= 0 and !visited[x - 1][y] and arr[x - 1][y] == '.')
        {
            visited[x - 1][y] = true;
            q.push({x - 1, y});
        }
        if (y + 1 < m and !visited[x][y + 1] and arr[x][y + 1] == '.')
        {
            visited[x][y + 1] = true;
            q.push({x, y + 1});
        }
        if (y - 1 >= 0 and !visited[x][y - 1] and arr[x][y - 1] == '.')
        {
            visited[x][y - 1] = true;
            q.push({x, y - 1});
        }
    }
    ++steps;
}

// for (auto x : time)
// {
//     for (auto y : x)
//         cout << y << " ";
//     cout << endl;
// }

for (ll i = 0; i < n; i++)
{
    for (ll j = 0; j < m; j++)
        visited[i][j] = false;
}

queue<pair<string, pair<ll, ll>>> pq;
pq.push({"", {b, c}});
visited[b][c] = true;
steps = 0;
while (!pq.empty())
{
    ll sz = pq.size();
    while (sz--)
    {
        auto front = pq.front();
        pq.pop();
        string res = front.first;
        ll x = front.second.first, y = front.second.second;
        // cout << x << " " << y << endl;
        visited[x][y] = true;

        if (x == 0 or x == n - 1 or y == 0 or y == m - 1)
        {
            // cout << x << " " << y << endl;
            cout << "YES" << endl;
            cout << steps << endl
                 << res;
            return 0;
        }

        if (x + 1 < n and !visited[x + 1][y] and arr[x + 1][y] == '.' and steps + 1 < time[x + 1][y])
        {
            visited[x + 1][y] = true;
            pq.push({res + 'D', {x + 1, y}});
        }
        if (x - 1 >= 0 and !visited[x - 1][y] and arr[x - 1][y] == '.' and steps + 1 < time[x - 1][y])
        {
            visited[x - 1][y] = true;
            pq.push({res + 'U', {x - 1, y}});
        }
        if (y + 1 < m and !visited[x][y + 1] and arr[x][y + 1] == '.' and steps + 1 < time[x][y + 1])
        {
            visited[x][y + 1] = true;
            pq.push({res + 'R', {x, y + 1}});
        }
        if (y - 1 >= 0 and !visited[x][y - 1] and arr[x][y - 1] == '.' and steps + 1 < time[x][y - 1])
        {
            visited[x][y - 1] = true;
            pq.push({res + 'L', {x, y - 1}});
        }
    }
    ++steps;
}
cout << "NO";

return 0;

    }

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      3 years ago, hide # ^ |
      ← Rev. 2  
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      your code is pushing too many strings into a queue, which each of them are O(n), where n is the length of the string, and that's what makes your code slow. Your approach is totally fine, I guess. So you just need to change the last part of the code.

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3 years ago, hide # |
 
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You should learn Multi source BFS

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