A.Shan and the Pirates.

Let us state the facts :

• We must go from the top left corner to the bottom right corner .

• We can only move right or down.

You may think that this question is dreadfully complicated as there are $$$^{m+n-2}C_{m-1}$$$ ways to go from the top left corner to the bottom right. If we were to check every possible path for the one with most treasures time taken by the program will be too much and we will clearly get TLE for larger m and n.

Let us take a coordinate system with $$$(1,1)$$$ being the top left corner of the grid and the bottom right being $$$(m,n)$$$. Define a function $$$f(x,y)$$$ to be the maximum number of treasures they can collect for all paths from $$$(1,1)$$$ to $$$(x,y)$$$ by only travelling down and right.

Take the input example in the question: \begin{array}{|c|c|c|c|} \hline 1 & 0 & 2 & 2 \cr \hline 0 & 0 & 0 & 0 \cr \hline 0 & 0 & 0 & 0 \cr \hline 0 & 3 & 0 & 1 \cr \hline \end{array}

Let $$$treasure(x,y)$$$ give the number of treasures on the island at $$$(x,y)$$$.

We can define the recursive relation : $$$f(x,y) = treasure(x,y) + max(f(x-1,y),f(x,y-1))$$$

This is because the we can reach $$$(x,y)$$$ immediately from the the points $$$(x-1,y)$$$ and $$$(x,y-1)$$$ only. If we had a $$$f(x-1,y)$$$ treasures at $$$(x-1,y)$$$ and we move right by one unit we will now have $$$f(x-1,y)+treasure(x,y)$$$.

Similarly ,If we had a $$$f(x,y-1)$$$ treasures at $$$(x,y-1)$$$ and we move down by one unit we will now have $$$f(x,y-1)+treasure(x,y)$$$.

The larger of these two values will $$$f(x,y)$$$ :-)

How do we find $$$f(m,n)$$$ -- the maximum number of treasures we can have by travelling from (1,1) to (m,n)?

Interestingly , we can just find the value of f for every point by iterating through every line one after another. I was initially proposing finding diagonals one by one , then realized line by line works just fine. Lets define $$$f(x,0)=0$$$ and $$$f(0,y)=0$$$ because there are no islands at those locations also, and it would make implementation of loop easier. If we did not define this , we would need to write separate loops to find value of f for first column and row.( its not too hard to do that but we're just lazy right).

The array of values of f are :

\begin{array}{|c|c|c|c|} \hline 1 & 1 & 3 & 5 \cr \hline 1 & 1 & 3 & 5 \cr \hline 1 & 1 & 3 & 5 \cr \hline 1 & 1 & 3 & 6 \cr \hline \end{array}

Now that the logical part is over , here is the code. Check out my video editorial too where I coded this !

Code in C++ :

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


int main(){
    int n,m;
    cin>>n>>m;
    int treasure[n][m];
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            cin>>treasure[i][j];
        }
    }
    
    //INPUT DONE
    
    int f[n+1][m+1]; //function , maximum treasures shan can have at an island (x,y)
    for(int i=0;i<=n;i++)f[i][0]=0;
    for(int i=0;i<=m;i++)f[0][i]=0;
    
    //set top , left side to ZERROOO
    
    for(int i=1;i<n+1;i++){
        for(int j=1;j<m+1;j++){
            f[i][j]=max(f[i-1][j],f[i][j-1])+treasure[i-1][j-1];
        }
    }
    
    cout<<f[n][m];
    
}