Single Round Match 626

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Автор Pheonix_0, 12 лет назад, По-английски

How to solve Single Round Match 626 Round 1 — Division I, Level Two NegativeGraphDiv1 Problem statement- http://community.topcoder.com/stat?c=problem_statement&pm=13218&rd=15859 thnxz in advance :)

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Pheonix_0
12 лет назад
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ASverdlov

12 лет назад, скрыть # |

← Rev. 3 →

I'll just translate Endagorion's solution he explained to me.

Calculate arrays zero[i][j] — minimum cost of path from vertex i to j if no edge is inverted, one[i][j] — minimum cost of path from vertex i to j if at most 1 edge is inverted.
Denote d[k][i][j] — minimum cost of path, where at most k edges were inverted. Then, d[k][i][j] = min(d[k — 1][i][u] + one[u][j]) for u = 1..n. And our answer stores in d[charges][1][n].
Notice that above formula is matrix multiplication d[k][i][j] * one[i][j] ((min, +) instead of usual (+, *)).

So our answer is zero[i][j] * ((one[i][j]) ^ charges) that can be compute in O(logN).

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Pheonix_0

12 лет назад, скрыть # |

Can you please elaborate your point 3.I didnt get it. thnxz

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ASverdlov

12 лет назад, скрыть # ^ |

← Rev. 2 →

I didn't get that part too:) To multiply two matrices, for every row of first matrix and column of second matrix you compute their scalar multiplication (scalarMult = A[row][1] * B[1][column] + A[row][2] * B[2][column] + ... + A[row][n] * B[n][column]). It turns out that we can change scalar multiplication to any operation with two arrays, such that the associative property saves. In our case we can apply this operation (change all "+" in scalar multiplication to "min", and all "*" to "+":

ourOperation = min(A[row][1] + B[1][column], A[row][2] + B[2][column], ... , A[row][n] + B[n][column]).

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