antivenom's blog

How can i construct a DP solution for this question????
By antivenom, 10 years ago, In English

Hello,After seeing this question the first thing came to my mind was Dynamic programming.I tried constructing a solution but could not make it,maybe because i don't have that much of experience.

  #include <bits/stdc++.h>
  using namespace std;
  int main() 
  {
      int n,m,a,b; 
      cin>>n>>m>>a>>b;
      int final=0,dp[n+1];
      dp[0]=0;
      for(int i=1;i<=n;i++)
	 dp[i]=min(dp[i-1]+a,dp[i-m]+b);
      cout<<dp[n];
      return 0;
 }

P.S:I know it can be solved without using Dp,but i request you to give Dp approach if it is possible.

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Comments Comments (8)
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10 years ago, # |
← Rev. 5   Vote: I like it 0 Vote: I do not like it

for i =  1 ... m - 1 dp[i - m] is undefined.

you may have WA in cases when prefered buy the special ticket for more than n rides.

correct solution

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10 years ago, # |
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if(i>=m) dp[i] = min(dp[i-1]+a,dp[i-m]+b); else dp[i]=min(dp[i-1]+a,b);

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    10 years ago, # ^ |
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    After putting your condition it fails on testcase 8.Before that it used to fail on 3.So some improvement is there.

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      10 years ago, # ^ |
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      My solution Got accepted lol

      Link Can you post a link of your solution ?

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        10 years ago, # ^ |
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        Bingo!!!Correct answer!!!...Please explain me else part of the recursion :)

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          10 years ago, # ^ |
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          With a ticket for 'm' rides, you can travel m times or less.

          Lets take this example : 7 5 5 1 Here the minimum cost for travelling 1...5 times is 1 . so DP[1...5] should have a value of 2.

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            10 years ago, # ^ |
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            can you explain the second test case 5 2 2 3 how the answer is 8?

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              10 years ago, # ^ |
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              Sure . Split 5 rides into 2 rides + 2 rides + 1 ride .
              The cost for 2 rides = 3 and the cost for 1 ride = 2.

              So you get 3+3+2 = 8

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