Help needed in ICPC Amritapuri 2010 problems

Revision en9, by bhishma, 2017-12-02 15:03:01

Recently we were solving problems from past Indian ICPC regional . We weren't able to solve these 2 problems , and I couldn't find any editorial either . It would be really helpful if you guys can give me some hints to these problems.

Chemicals [Solved]


There are N bottles each having a different chemical. For each chemical i, you have determined C[i], which means that mixing chemicals i and C[i] causes an explosion. You have K distinct boxes. In how many ways can you divide the N chemicals into those boxes such that no two chemicals in the same box can cause an explosion together?


  • T ≤ 50
  • 2 ≤ N ≤ 100
  • 2 ≤ K ≤ 1000

My Ideas

I thought of modelling the given dependencies as a graph and we are asked to find the number of ways to partition the graph into independent sets . But I realised that counting independent sets is intractable , so there must be a much more efficient or different way to solve this problem .


Thanks to ideas from fram and AleksanderBalobanov Since it's a functional graph we will have exactly one cycle per connected component . We need to color the graph using atmost K colors such that no 2 adjacent vertices have the same color . For any node which is not a cycle we have K - 1 options . For the nodes in a cycle we need to take care of them separately . We can construct a recurrence to find the number of ways to color a cycle in the following way.

Let DP[i] denote the number of ways to color a chain of length i such that it starts and ends with different colors.

DP[1] = 0, DP[2] = K * (K - 1)

DP[i] = (K - 2) * DP[i - 1] + (K - 1) * DP[i - 2]

This is basically saying that we have (K - 2) options for a chain starting and ending with different colors and (K - 1) options for a chain starting and ending with the same color. Another observation is that the number of chains of length i - 1 which starts and ends with same color is same as number of chains of length i - 2 which starts and ends with different colors.

My solution

Dividing Stones [Solved]


There are N stones, which can be divided into some piles arbitrarily. Let the value of each division be equal to the product of the number of stones in all the piles modulo P. How many possible distinct values are possible for a given N and P?


  • T ≤ 20
  • 2 ≤ N ≤ 70
  • 2 ≤ P ≤ 109

Idea (thanks to AleksanderBalobanov)

We represent every partition as p1a1 * p2a2 * ... * pkak, where p1, p2, ..., pk are primes up to 70. We can achieve this value for a given n iff a1 * p1 + a2 * p2 + ... + ak * pk ≤ n . So the partition looks like p1 + p1...a1 times + ... + pk + pk...ak times , if the value is less than n we can add extra 1 .

My solution

Tags acm-icpc amritapuri, graph, math


  Rev. Lang. By When Δ Comment
en9 English bhishma 2017-12-02 15:03:01 0 Solved Chemicals (published)
en8 English bhishma 2017-12-02 15:01:05 14
en7 English bhishma 2017-12-02 14:59:17 1209 (saved to drafts)
en6 English bhishma 2017-11-11 08:04:33 8 solved dividing stones (published)
en5 English bhishma 2017-11-11 08:02:39 513 (saved to drafts)
en4 English bhishma 2017-09-08 21:37:57 38 (published)
en3 English bhishma 2017-09-08 20:55:08 483
en2 English bhishma 2017-09-08 20:05:56 235
en1 English bhishma 2017-09-08 19:55:40 958 Initial revision (saved to drafts)