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cgy4ever's blog

Topcoder SRM 707

By cgy4ever, history, 8 years ago, In English

Topcoder SRM 707

Time: 9:00 pm EST Monday, January 30, 2017

Calendar: https://www.topcoder.com/community/events/

cgy4ever
8 years ago
53

Comments (48)

Show archived | Write comment?

cgy4ever

8 years ago, # |

Update: we added one more SRM on Jan 11th and moved the SRM on 14th to 21st.

So we will have 3 SRMs in January! See details: 705, 706, 707

→ Reply

cgy4ever

8 years ago, # |

+41

SRM 707 will start in 24 hours!

→ Reply

animeshf

8 years ago, # |

Could people describe some clean solutions for Div 1 250?

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cegprakash

8 years ago, # ^ |

I was thinking in terms of BFS initially couldn't get any clue.. Then tried zig zag patterns. No idea at all :(

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Morphy

8 years ago, # ^ |

Is always possible to generate a pattern of the following form:

→ Reply

animeshf

8 years ago, # ^ |

Thanks! I guess this can be done in O(N²) too, but O(N⁴) seems safer to code for TopCoder SRMs.

→ Reply

cegprakash

8 years ago, # ^ |

This was my intuition. But how to find the rows and cols required and the zig zag pattern?

Note that the above pattern does not have any up direction moves.

We should also consider the symmetric case where we have to do a pattern with no left direction moves.

→ Reply

animeshf

8 years ago, # ^ |

Take a 50 * 50 grid with all cells "Safe". Note that the initial shortest path is 98. Handle K < 98 separately.

Then, whenever you are forced to move one step left/up, you increase the shortest path by 2. Thus, if you block enough cells you can make any even number in the range [100, 1000].

For odd K, use a 49 * 50 grid and repeat the same process.

→ Reply

cegprakash

8 years ago, # ^ |

I don't understand. How do u decide where to place the walls?

→ Reply

animeshf

8 years ago, # ^ |

Suppose your grid is n * m. You place it in the following order :-

(0, 1) (1, 1) (2, 1) ... (n - 2, 1)

(n - 1, 3) (n - 2, 3) (n - 3, 3) ... (1, 3)

(0, 5) (1, 5) (2, 5) ... (n - 2, 5)

and so on....

Basically you're forcing an up movement here and constraints are small enough for you to compute the current shortest path after each addition.

→ Reply

george24601

8 years ago, # ^ |

How to prove that this zig-zag approach gives the maximum k, i.e., it gives the maximum number of left steps + up steps?

→ Reply

animeshf

8 years ago, # ^ |

What do you mean by maximum number? Restating what I said earlier, every "up" movement that you make increases your shortest path by 2, so if K is even and n + m - 2 is even, then you will reach K by forcing sufficient number of "up" movements. The case where K is odd is analogous, just make sure that n + m - 2 is odd.

→ Reply

animeshf

8 years ago, # ^ |

Here is some code for reference.

→ Reply

george24601

8 years ago, # ^ |

← Rev. 2 →

Ignore this comment

Thanks for the reply.

"if you block enough cells you can make any even number in the range [100, 1000]."

Say for a 50 * 50 grid, how to prove that 1000 is the upper bound, and not something higher?

In other words, how to prove max number of "up" movements we can force on a board?

→ Reply

KADR

8 years ago, # ^ |

It doesn't. For instance, there's a pattern which can give up to $\text{[math]}$ steps but it's too complicated for 250.

→ Reply

george24601

8 years ago, # ^ |

Ah I see why I was confused, I didn't pay attention to k <= 1000 constraint. Silly me!

Thanks a lot for your help!

→ Reply

King_George

8 years ago, # ^ |

-6

Could you please share with us how to make that pattern?

→ Reply

Chelly

8 years ago, # ^ |

← Rev. 2 →

+26

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SarvagyaAgarwal

8 years ago, # |

Solution for Div2 500 and 1000 ? everybody failed Div2 500 .

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antonio.deluna

8 years ago, # ^ |

There must be a mistake in the system test for Div2 500

→ Reply

SarvagyaAgarwal

8 years ago, # ^ |

Ratings are already updated though .

→ Reply

yassin_

8 years ago, # ^ |

Look at the systest, test 121 is wrong, I'm sure

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Hiasat

8 years ago, # |

← Rev. 2 →

+22

There is an invalid test case in Div2-500

{{"...", ".#.", "..#"}, 4}

Expected (System test output): "DDRR"

Received (My Output): ""

Answer checking result: There exist a solution, but your output is "".

→ Reply

cgy4ever

8 years ago, # ^ |

← Rev. 2 →

Yes, we are fixing it.

Sorry for the delay.

Update: Fixed now.

→ Reply

CodeItLikeAidos

8 years ago, # ^ |

Return "Exists" if there is at least one cross on the given board. Otherwise, return "Does not exist". Note that the return value is case-sensitive.

Exist — Exists. I couldn't find bug around 30 minutes. System expected "Exist".

→ Reply

AM51

8 years ago, # ^ |

Please check my solution (Handle — AM51) . System tests are passing but someone challenged the solution before . Are you also reconsidering the challenges that might have been made with wrong cases ?

→ Reply

P___

8 years ago, # ^ |

How did it affect challenge phase in div2?

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zscoder

8 years ago, # |

What's the correct way to solve Div 1 450?

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abcdef6199

8 years ago, # ^ |

← Rev. 2 →

After applying K multiply operations and L add operations, S can be express in the form:

S * B^K + A(B^p1 + B^p2 + ... + B^pL) where pi < = K

Iterate K from 0 to 64 (or a bit bigger), consider remain = (T - S * B^K) / A, we'll try to express this as sum of power of B. This can be done with a simple greedy.

And make sure to check the case B = 0 first. I failed because of this :(

→ Reply

zscoder

8 years ago, # ^ |

I actually checked the case B = 0 wrongly. I didn't realize you can jump back to 0 if B = 0 -_-

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Shadek

8 years ago, # ^ |

← Rev. 2 →

Is there any extra constraint like

p1 >= p2 >= .. >= PL=1

if we consider p1 is largest among them.

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abcdef6199

8 years ago, # ^ |

It doesn't matter. Just 0 ≤ P_i ≤ K will do.

→ Reply

SarvagyaAgarwal

8 years ago, # |

approach for Div2 1000 ?

→ Reply

egor.okhterov

8 years ago, # |

← Rev. 2 →

Can anyone help me understand one thing.
How does the following code may result in segmentation fault for the following input:
{{".#...", ".#.#.", ".#.#.", ".#.#.", "...#."}, 3000}

Div2 500

#include <iostream>
#include <string>
#include <vector>

int n, m;
bool cell[50][50];
char p[3000][50][50];

int dr[4] = { 1, 0, -1, 0 };
int dc[4] = { 0, -1, 0, 1 };
char dir[4] = { 'D', 'L', 'U', 'R' };

bool valid(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m && cell[r][c]; }

void dfs(int r0, int c0, int steps)
{
    if (steps > 3000 || !valid(r0, c0)) return;

    for (int i = 0; i < 4; i++)
    {
        int r = r0 + dr[i];
        int c = c0 + dc[i];

        if (valid(r, c) && !p[steps][r][c])
        {
            p[steps][r][c] = dir[i];
            dfs(r, c, steps + 1);
        }
    }
}

struct StepsConstruct
{
    static string construct(vector <string> board, int k);
};

string StepsConstruct::construct(vector<string> board, int k)
{
    n = board.size();
    m = board[0].size();
    for (int r = 0; r < n; r++)
        for (int c = 0; c < m; c++)
            cell[r][c] = board[r][c] == '.';

    dfs(0, 0, 1);

    int r = n - 1;
    int c = m - 1;
    if (!p[k][r][c]) return "";

    int id[256];
    id['U'] = 0;
    id['D'] = 2;
    id['L'] = 3;
    id['R'] = 1;

    string ans;
    while (k > 0)
    {
        int dd = p[k][r][c];

        ans.push_back(dd);
        r += dr[id[dd]];
        c += dc[id[dd]];
        k--;
    }

    reverse(ans.begin(), ans.end());
    return ans;
}

I cannot reproduce that problem locally and I get the correct output.

→ Reply

CodeItLikeAidos

8 years ago, # ^ |

array p has maximal index 2999 and steps variable can be 3000.

→ Reply

egor.okhterov

8 years ago, # ^ |

Thank you, now passed all systests ;(

→ Reply

pakhandi

8 years ago, # |

← Rev. 2 →

+15

Screencast-HackPhase

→ Reply

selfcompiler

8 years ago, # |

Can anyone please explain DIV 2 500 problem ?

→ Reply

SarvagyaAgarwal

8 years ago, # ^ |

The way i solved it : Write a recursive function bool canSolve(int row,int col,int stepsLeft) which tells you if it is possible to reach cell (n-1,m-1) from cell (row,col) in exactly stepsLeft steps . Transitions are pretty straightforward , just recurse on all 4 adjacent cells .

→ Reply

selfcompiler

8 years ago, # ^ |

where to stop that recursion when steps left becomes zero ?

→ Reply

SarvagyaAgarwal

8 years ago, # ^ |

Base Case : if(stepsLeft == 0) { return row == n-1 && col == m-1 ; }

→ Reply

selfcompiler

8 years ago, # ^ |

← Rev. 3 →

can we have this approach 1. find the shortest path such that ((K-pathlength)==even_number) 2. print that path 3. repeat last two consecutive characters (K-pathlength) times .

→ Reply