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<style> .ss {font-family:sans-serif} </style>
Nintendo's new minimalist video game controller only has two buttons,
down
and
right.
In the platform launch title Super Mario Multiverse,
every level is a square grid of cells
where Mario starts in the top-left corner. Some
cells of the grid have coins with a value, and other cells
are empty. The player has to press down
and right in some order until
they get to the bottom-right corner, and their score is the sum total of
the value of
all coins they hit along the way. Write a program
to find the maximum possible score for any level.


We represent a level by an integer <tt>n</tt> giving the square room size,
then an <tt>n</tt>-by-<tt>n</tt> grid <tt>cellValue</tt> where
<tt>cellValue[i][j]</tt> is the cell <tt>i</tt>th from the top
and <tt>j</tt>th from the left.
A positive value means a coin with that value,
and a value of 0 means no coin is in that cell.


For example, suppose the input is
<pre>
5
1 1 0 2 0
0 2 0 2 0
0 1 0 0 1
3 0 0 0 1
0 0 0 0 0
</pre>
Mario has lots of possible paths. Remember he must go from the top-left
to the bottom-right using only steps right
(R) and down (D).
If he goes D-D-D-R-R-R-R-D, he gets a score of 1+3+1=5. The path
R-D-R-R-D-R-D-D achieves a score of 1+1+2+2+1+1=8, which happens to be
the maximum possible for this room.

Write a function <code>int maxPath(int n, int** cellValue)</code>
that computes the maximum score for a level given in this format.
Use dynamic programming: compute, for each position, the best value
you can get if you start in that cell.

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Choices
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Correct answer
Template / Reference solution

Template / Reference solution

#include <iostream>
#include <algorithm>
using namespace std;

int maxPath(int n, int** cellValue) {
// opt[i][j]: max value attainable if starting at position [i][j]
int opt[n][n]; // bad style!

// base case: corner
opt[n-1][n-1] = cellValue[n-1][n-1];
// pseudo-base case: right column, have to go down
for (int i=n-2; i>=0; i--)
opt[i][n-1] = cellValue[i][n-1] + opt[i+1][n-1];
// pseudo-base case: bottom row, have to go right
for (int j\[=n-2; j>=0; j--]\)
\[opt[n-1][j] = cellValue[n-1][j] + opt[n-1][j+1]]\;
// general case
for (int i\[=n-2; i>=0; i--]\) {
for (int j\[=n-2; j>=0; j--]\) {
// a choice! take better choice of down and right
opt[i][j] = cellValue[i][j] + \[max]\(\[opt[i+1][j]]\,
\[opt[i][j+1]]\);
}
}
// value in top-left corner
return \[opt[0][0]]\;
}

int main() {
int n;
cin >> n;
int** coins = new int*[n];
for (int i=0; i<n; i++) coins[i] = new int[n];
for (int i=0; i<n; i++)
for (int j=0; j<n; j++)
cin >> coins[i][j];
cout << maxPath(n, coins);
}

Java test suite
See manual

C++ test suite
json list of stdin/args tests
e.g. [{"stdin":"hi", "args":["4", "5"]}, {"stdin":"noargs"}]
to just run once with no input use [{}]

[
{"stdin": "5\n1 1 0 2 0\n0 2 0 2 0\n0 1 0 0 1\n3 0 0 0 1\n0 0 0 0 0"},
{"stdin": "3\n31 2 0\n5 26 6\n0 4 33"},
{"stdin": "3\n0 2 31\n6 26 5\n33 4 0"},
{"stdin": "4\n5 3 1 0\n0 3 4 0\n5 3 4 0\n0 2 1 0"},
{"stdin": "2\n1 2\n3 4"},
{"stdin": "10\n1 5 8 4 1 0 2 4 0 6\n8 3 6 8 6 4 6 2 8 0\n3 6 0 1 1 9 8 4 2 0\n2 0 4 4 1 9 1 3 3 3\n2 0 4 8 9 6 5 0 8 2\n9 5 7 6 0 0 5 4 4 3\n7 4 9 1 4 4 1 4 8 2\n6 1 4 7 0 8 5 8 3 8\n9 6 0 2 3 6 6 4 5 2\n5 0 5 1 6 1 4 3 5 0"}
]

C++ test suite
See manual

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Description html markup shown to student	<style> .ss {font-family:sans-serif} </style> Nintendo's new minimalist video game controller only has two buttons, <span class='ss'>down</span> and <span class='ss'>right</span>. In the platform launch title Super Mario Multiverse, every level is a square grid of cells where Mario starts in the top-left corner. Some cells of the grid have coins with a value, and other cells are empty. The player has to press <span class='ss'>down</span> and <span class='ss'>right</span> in some order until they get to the bottom-right corner, and their score is the sum total of the value of all coins they hit along the way. Write a program to find the maximum possible score for any level. <p> We represent a level by an integer <tt>n</tt> giving the square room size, then an <tt>n</tt>-by-<tt>n</tt> grid <tt>cellValue</tt> where <tt>cellValue[i][j]</tt> is the cell <tt>i</tt>th from the top and <tt>j</tt>th from the left. A positive value means a coin with that value, and a value of 0 means no coin is in that cell. <p> For example, suppose the input is <pre> 5 1 1 0 2 0 0 2 0 2 0 0 1 0 0 1 3 0 0 0 1 0 0 0 0 0 </pre> Mario has lots of possible paths. Remember he must go from the top-left to the bottom-right using only steps <span class='ss'>right</span> (R) and <span class='ss'>down</span> (D). If he goes D-D-D-R-R-R-R-D, he gets a score of 1+3+1=5. The path R-D-R-R-D-R-D-D achieves a score of 1+1+2+2+1+1=8, which happens to be the maximum possible for this room. <p> Write a function <code>int maxPath(int n, int** cellValue)</code> that computes the maximum score for a level given in this format. Use dynamic programming: compute, for each position, the best value you can get if you start in that cell.
Epilogue html markup shown when solved	remove
Public permissions	remove
Remarks Comments, history, license, etc.	remove
Engine
Choices json list of string, bool pairs e.g. `[["good", true], ["bad", false]]`
Correct answer
Template / Reference solution
Template / Reference solution	#include <iostream> #include <algorithm> using namespace std; int maxPath(int n, int cellValue) { // opt[i][j]: max value attainable if starting at position [i][j] int opt[n][n]; // bad style! // base case: corner opt[n-1][n-1] = cellValue[n-1][n-1]; // pseudo-base case: right column, have to go down for (int i=n-2; i>=0; i--) opt[i][n-1] = cellValue[i][n-1] + opt[i+1][n-1]; // pseudo-base case: bottom row, have to go right for (int j\[=n-2; j>=0; j--]\) \[opt[n-1][j] = cellValue[n-1][j] + opt[n-1][j+1]]\; // general case for (int i\[=n-2; i>=0; i--]\) { for (int j\[=n-2; j>=0; j--]\) { // a choice! take better choice of down and right opt[i][j] = cellValue[i][j] + \[max]\(\[opt[i+1][j]]\, \[opt[i][j+1]]\); } } // value in top-left corner return \[opt[0][0]]\; } int main() { int n; cin >> n; int coins = new int*[n]; for (int i=0; i<n; i++) coins[i] = new int[n]; for (int i=0; i<n; i++) for (int j=0; j<n; j++) cin >> coins[i][j]; cout << maxPath(n, coins); }
Java test suite See manual
C++ test suite json list of stdin/args tests e.g. `[{"stdin":"hi", "args":["4", "5"]}, {"stdin":"noargs"}]` to just run once with no input use `[{}]`	[ {"stdin": "5\n1 1 0 2 0\n0 2 0 2 0\n0 1 0 0 1\n3 0 0 0 1\n0 0 0 0 0"}, {"stdin": "3\n31 2 0\n5 26 6\n0 4 33"}, {"stdin": "3\n0 2 31\n6 26 5\n33 4 0"}, {"stdin": "4\n5 3 1 0\n0 3 4 0\n5 3 4 0\n0 2 1 0"}, {"stdin": "2\n1 2\n3 4"}, {"stdin": "10\n1 5 8 4 1 0 2 4 0 6\n8 3 6 8 6 4 6 2 8 0\n3 6 0 1 1 9 8 4 2 0\n2 0 4 4 1 9 1 3 3 3\n2 0 4 8 9 6 5 0 8 2\n9 5 7 6 0 0 5 4 4 3\n7 4 9 1 4 4 1 4 8 2\n6 1 4 7 0 8 5 8 3 8\n9 6 0 2 3 6 6 4 5 2\n5 0 5 1 6 1 4 3 5 0"} ]
C++ test suite See manual
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Remove default compiler flag(s) json list of flags see default_cppflags e.g. `["-Wno-write-strings"]`	remove

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Note: problems are open-source by default (see 'Public permissions'). Assumed license for open problems is Creative Commons 4.0 Attribution-ShareAlike unless specified in 'Remarks'.