20 lines
1.3 KiB
Markdown
20 lines
1.3 KiB
Markdown
|
As the problem description states this is similar to problem-081.
|
||
|
|
|
||
|
|
Again we generate an node-array where we remark the minimal distance from the left side to this node.
|
||
|
|
Initially we can initialize the left column with its input values.
|
||
|
|
(`distance[0, y] = data[0, y]`) and all the other nodes with an absurdly high number.
|
||
|
|
|
||
|
|
Then we iterate through all remaining columns:
|
||
|
|
|
||
|
|
For each column `x` we go all possible ways from the previous column. That means:
|
||
|
|
- Choose the start-row `y` (and do this for all possible start rows)
|
||
|
|
- Get the distance to reach this row by calculating `distance[x-1, y] + data[x, y]`
|
||
|
|
- Then go all the way up and down and calculate the distance on the way `distance[x-1, y] + data[x, y] + data[x, y - 1] + data[x, y - 2] ...`
|
||
|
|
- For each node where this distance is lesser than the current one we update distance array.
|
||
|
|
- *Optimization node:* Once we find a node where the distance is greater than a previous calculated we can stop further traversing the column (in this direction)
|
||
|
|
|
||
|
|
At the end we have an distance array where each node is the minimal distance to reach this node from the left side.
|
||
|
|
Our result is then the minimal value of the most-right column.
|
||
|
|
|
||
|
|
*Note:* While problem-081 hat an time complexity of O(n) this one has one of O(n^2).
|
||
|
|
But for an 80x80 array that's still fast enough and really not an problem.
|