# Algorithm practice - Treasure Island 1

## Description

You have a map that marks the location of a treasure island.

Some of the map area has jagged rocks and dangerous reefs. Other areas are safe to sail in.

There are other explorers trying to find the treasure. So you must figure out a shortest route to the treasure island.

Assume the map area is a two dimensional grid, represented by a matrix of characters.

You must start from the top-left corner of the map and can move one block up, down, left or right at a time.

The treasure island is marked as ‘X’ in a block of the matrix. ‘X’ will not be at the top-left corner.

Any block with dangerous rocks or reefs will be marked as ‘D’. You must not enter dangerous blocks. You cannot leave the map area.

Other areas ‘O’ are safe to sail in. The top-left corner is always safe.

Output the minimum number of steps to get to the treasure.

e.g.
Input
[
[‘O’, ‘O’, ‘O’, ‘O’],
[‘D’, ‘O’, ‘D’, ‘O’],
[‘O’, ‘O’, ‘O’, ‘O’],
[‘X’, ‘D’, ‘D’, ‘O’],
]

Route is (0, 0), (0, 1), (1, 1), (2, 1), (2, 0), (3, 0)
The minimum route takes 5 steps.


## Solution

import sys

def solution(treasure_map):
# Start from left corner
start_x = len(treasure_map)
start_y = len(treasure_map[0])
treasure_x = 0
treasure_y = 0
distance_map = []
for row in treasure_map:
distance_row = []
for col in row:
if col == 'O':
distance_row.append(1)
elif col == 'D':
distance_row.append(1000000)  # Big enough
elif col == 'X':
distance_row.append(1)
distance_map.append(distance_row)
distance_map[0][0] = 0
'''
for all position:
cursor = current_position
if cursor is D:
pass
cur_distance_from_upper_left = min(from_left_up, from_up, from_right_up, from_left)
return X
'''
for x in range(start_x):
for y in range(start_y):
cursor = treasure_map[x][y]
if cursor == 'D':
continue
elif cursor == 'X':
treasure_x = x
treasure_y = y
else:
candidate = []
if x != 0:  # up pass
candidate.append(distance_map[x - 1][y])  # from_up
if y != 0:
candidate.append(distance_map[x][y - 1])  # from_left
if len(candidate) != 0 and candidate != 'D':
distance_map[x][y] = min(candidate) + distance_map[x][y]
return distance_map[treasure_x][treasure_y]

if __name__ == "__main__":
input_map = []
while True:
if l == ']':
break
elif l == '[':
continue
li = l.split(',')
input_map.append(list(
filter(None, [str(x.replace('[', '').replace(']', '').replace("'", "").strip()) for x in li])))
print(input_map)
print(solution(input_map))

'''
[
['O', 'O', 'O', 'O'],
['D', 'O', 'D', 'O'],
['O', 'O', 'O', 'O'],
['X', 'D', 'D', 'O'],
]
'''



## Short Description

• Approach
• I took some time to parse input and strip for handy data.
• I need to shortest path for ‘X’ sign.
• When the loop found solution, we need to keep at moment to find whether there is another solution more closer to duration.
• If all loop are traversed, the final list is solution.

태그:

카테고리:

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