DAA Implementation Project

/*
DAA - IMPLEMENTATION PROJECT
Done By: IMT2019525 VIJAY JAISANKAR, IMT2019049 MADDULA DHANUSH
*/

#include<bits/stdc++.h>
using namespace std;


/*
   This is the dp table.
   At the end of the program,dp[i][j] will contain the minimum cost of merging the slimes in the segment i,...,j
*/
long long int dp[401][401];



/*
   This is the array containing partial sum/prefix sum values.
   After preprocessing, pfsum[i] = arr[0] + arr[1] + ... + arr[i],
   (where i is the input array)
*/
long long int pfsum[401];



/*
   This is the input array containing the slime values.
*/
long long int arr[401];



/*
   A dummy variable used to store the theoretical maximum value possible in this context
*/
long long int INF = 1LL<<60;



int main(){

    /*
       n is the size of the input.
    */
    int n;
    cin>>n;



    /*
       Constructing the prefix sum array while taking the input:
       We have desctibed this process in the paper - note that this whole process is O(n)
    */
    for(int i=0;i<n;i++){
        cin>>arr[i];
        if(i==0)    pfsum[i] = arr[i];
        else        pfsum[i] = pfsum[i-1] + arr[i];
    }



    /*
       Setting up base cases and initialising the dp[][] array:
       Base cases: dp[i][i] = 0 - this is to facilitate the vacuous step of the recurrence
       Otherwise, as we are going to minimize the dp values, we set them to the aforementioned INF
    */
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(i==j)    dp[i][j] = 0;
            else        dp[i][j] = INF;
        }
    }



    /*
       Implementing the recurrence in a bottom-up fashion:
       From the dependency graph, this is a way to build the DP in a bottom-up way.
       We can clearly see the recurrence in action while setting dp[l][r]
       Note that, as we need the non-recurisive merging work while considering every eligible value of k, we pre-store it.
    */
    for(int l=n-1;l>=0;l--){
        for(int r=l+1;r<n;r++){
            long long int pfdiff = pfsum[r] - pfsum[l-1];
            for(int k=l;k<r;k++){
                dp[l][r] = min(dp[l][r],dp[l][k] + dp[k+1][r] + pfdiff); // Testing out candidate values of k
            }
        }
    }



    /*
       Outputting the value:
       From the definition of dp[i][j], we need to print out dp[0][n-1] as we follow 0-based indexing.
    */
    cout<<dp[0][n-1]<<"\n";
    return 0;
}