CCC '24 J5 - Harvest Waterloo - DMOJ: Modern Online Judge

CCC '24 J5 - Harvest Waterloo

Points: 7 (partial)

Time limit: 3.0s

Memory limit: 1G

Problem types

Canadian Computing Competition: 2024 Stage 1, Junior #5

There is a wildly popular new harvest simulation game called Harvest Waterloo. The game is played on a rectangular pumpkin patch which contains bales of hay and pumpkins of different sizes. To begin the game, a farmer is placed at the location of a pumpkin.

The farmer harvests all pumpkins they can reach by moving left, right, up, and down throughout the patch. The farmer cannot move diagonally. The farmer can also not move through a bale of hay nor move outside of the patch.

Your job is to determine the total value of all the pumpkins harvested by the farmer. A small pumpkin is worth $\$1$ ~\$1~, a medium pumpkin is worth $\$5$ ~\$5~, and a large pumpkin is worth $\$10$ ~\$10~ dollars.

Input Specification

The first line of input is an integer $R > 0$ ~R > 0~ which is the number of rows within the patch.

The second line of input is an integer $C > 0$ ~C > 0~ which is the number of columns within the patch.

The next $R$ ~R~ lines describe the patch. Each line will contain $C$ ~C~ characters and each character will either represent a pumpkin size or a bale of hay: S for a small pumpkin, M for a medium pumpkin, L for a large pumpkin, or * for a bale of hay.

The next line of input is an integer $A$ ~A~ where $0 \le A < R$ ~0 \le A < R~, and the last line of input is an integer $B$ ~B~ where $0 \le B < C$ ~0 \le B < C~. Row $A$ ~A~ and column $B$ ~B~ is the starting location of the farmer and the top-left corner of the patch is row $0$ ~0~ and column $0$ ~0~.

The following table shows how the available 15 marks are distributed:

Marks	Description	Bound
1	The patch is small and there are no bales of hay.	$R \times C \le 100$ ~R \times C \le 100~
4	The patch is small and the bales of hay divide the entire patch into rectangular areas.	$R \times C \le 100$ ~R \times C \le 100~
5	The patch is small and the bales of hay can be anywhere.	$R \times C \le 100$ ~R \times C \le 100~
5	The patch is large and the bales of hay can be anywhere.	$R \times C \le 100 \,000$ ~R \times C \le 100 \,000~

Output Specification

Output the integer, $V$ ~V~, which is the total value in dollars of all the pumpkins harvested by the farmer.

Sample Input 1

6
6
**LMLS
S*LMMS
S*SMSM
******
LLM*MS
SSL*SS
5
1

Output for Sample Input 1

Explanation of Output for Sample Input 1

Starting at row $5$ ~5~ and column $1$ ~1~, the farmer can reach the $6$ ~6~ pumpkins in the highlighted area. They harvest $2$ ~2~ small pumpkins, $1$ ~1~ medium pumpkin, and $3$ ~3~ large pumpkins. The total value in dollars of this harvest is $2 \times 1 + 1 \times 5 + 3 \times 10 = 37$ .

Sample Input 2

6
6
**LMLS
S*LMMS
S*SMSM
***SLL
LLM*MS
SSL*SS
2
4

Output for Sample Input 2

Explanation of Output for Sample Input 2

Starting at row $2$ ~2~ and column $4$ ~4~, the farmer can reach the $19$ ~19~ pumpkins in the highlighted area. They harvest $8$ ~8~ small pumpkins, $6$ ~6~ medium pumpkin, and $5$ ~5~ large pumpkins. The total value in dollars of this harvest is $8 \times 1 + 6 \times 5 + 5 \times 10 = 88$ .

Comments

0

dhruvkangaroo commented on Feb. 5, 2025, 11:47 p.m.

remember if ur using python to not forget about sys.setrecursionlimit(x) you will get IR error if u dont

0

Have_A_Nice_Life commented on Feb. 7, 2025, 2:56 p.m.

or use bfs

0

ZakichanDraws commented on Feb. 6, 2025, 6:01 p.m.

how big do you have to set it

0

bowen_wu commented on Feb. 17, 2025, 12:50 p.m.

set it to 10^5 because the worst case is no haybales and r*c = 100000, in which case does 10^5 recursions. (note: maybe set it to 10^5 + 3 to prevent RecursionError because it works for me)

-2

dhruvkangaroo commented on Feb. 11, 2025, 3:25 p.m.

just set it to a gigantic number till it doesn't give one. sys.setrecursionlimit(10^7)

3

iwantbred commented on Jan. 25, 2025, 7:04 p.m.

happy early birthday