Topcoder SRM 707 - Codeforces

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Topcoder SRM 707

Автор cgy4ever, история, 9 лет назад, По-английски

Topcoder SRM 707

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

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

cgy4ever
9 лет назад
53

Комментарии (48)

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cgy4ever

9 лет назад, скрыть # |

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

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cgy4ever

9 лет назад, скрыть # |

+41

SRM 707 will start in 24 hours!

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animeshf

9 лет назад, скрыть # |

Could people describe some clean solutions for Div 1 250?

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cegprakash

9 лет назад, скрыть # ^ |

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

9 лет назад, скрыть # ^ |

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

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animeshf

9 лет назад, скрыть # ^ |

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

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cegprakash

9 лет назад, скрыть # ^ |

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.

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animeshf

9 лет назад, скрыть # ^ |

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.

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cegprakash

9 лет назад, скрыть # ^ |

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

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animeshf

9 лет назад, скрыть # ^ |

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.

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george24601

9 лет назад, скрыть # ^ |

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

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animeshf

9 лет назад, скрыть # ^ |

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.

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animeshf

9 лет назад, скрыть # ^ |

Here is some code for reference.

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george24601

9 лет назад, скрыть # ^ |

← 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?

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KADR

9 лет назад, скрыть # ^ |

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

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george24601

9 лет назад, скрыть # ^ |

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

Thanks a lot for your help!

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King_George

9 лет назад, скрыть # ^ |

-6

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

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Chelly

9 лет назад, скрыть # ^ |

← Rev. 2 →

+26

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SarvagyaAgarwal

9 лет назад, скрыть # |

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

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antonio.deluna

9 лет назад, скрыть # ^ |

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

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SarvagyaAgarwal

9 лет назад, скрыть # ^ |

Ratings are already updated though .

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yassin_

9 лет назад, скрыть # ^ |

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

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Hiasat

9 лет назад, скрыть # |

← 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 "".

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cgy4ever

9 лет назад, скрыть # ^ |

← Rev. 2 →

Yes, we are fixing it.

Sorry for the delay.

Update: Fixed now.

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CodeItLikeAidos

9 лет назад, скрыть # ^ |

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".

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AM51

9 лет назад, скрыть # ^ |

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 ?

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P___

9 лет назад, скрыть # ^ |

How did it affect challenge phase in div2?

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zscoder

9 лет назад, скрыть # |

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

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abcdef6199

9 лет назад, скрыть # ^ |

← 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 :(

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zscoder

9 лет назад, скрыть # ^ |

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

9 лет назад, скрыть # ^ |

← 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

9 лет назад, скрыть # ^ |

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

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SarvagyaAgarwal

9 лет назад, скрыть # |

approach for Div2 1000 ?

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egor.okhterov

9 лет назад, скрыть # |

← 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.

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CodeItLikeAidos

9 лет назад, скрыть # ^ |

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

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egor.okhterov

9 лет назад, скрыть # ^ |

Thank you, now passed all systests ;(

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pakhandi

9 лет назад, скрыть # |

← Rev. 2 →

+15

Screencast-HackPhase

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selfcompiler

9 лет назад, скрыть # |

Can anyone please explain DIV 2 500 problem ?

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SarvagyaAgarwal

9 лет назад, скрыть # ^ |

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 .

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selfcompiler

9 лет назад, скрыть # ^ |

where to stop that recursion when steps left becomes zero ?

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SarvagyaAgarwal

9 лет назад, скрыть # ^ |

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

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selfcompiler

9 лет назад, скрыть # ^ |

← 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 .

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