62 lines
1.9 KiB
Chapel
62 lines
1.9 KiB
Chapel
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use IO;
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// It's easiest to just yield all numbers as a list, then reshape that list
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// into a square given `rowSize` information.
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iter allNumbers(ref rowSize: int) {
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for line in stdin.lines() {
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const trimmedLine = line.strip();
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rowSize = trimmedLine.size;
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for num in trimmedLine {
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yield num : int;
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}
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}
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}
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// Time to do that reshaping!
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var rowSize = 0;
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var allNums = allNumbers(rowSize);
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var shapedNums = reshape(allNums, { 1..allNums.size / rowSize, 1..rowSize });
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// In this problem, we're considering the view from each direction at each
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// tree. Define a helper iterator to return an array slice representing trees
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// in that direction, as well as the "by" step indicating which direction
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// the path should be walked in.
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iter eachDirection((x,y)) {
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yield (shapedNums[..<x,y], -1);
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yield (shapedNums[x+1..,y], 1);
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yield (shapedNums[x,..<y], -1);
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yield (shapedNums[x,y+1..], 1);
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}
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// All that's left is to write special-case behavior for each part.
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// In part 1, we check if a tree is _never_ blocked along a direction; this can be
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// accomplished by checking whether or not all trees are less tall than the current tree.
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proc int.visibleAlong((trees, step)) {
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return max reduce trees < this;
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}
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// In part 2, we count the number of trees until a taller tree is encountered.
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// Here we iterate serially, and use our `step` parameter.
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proc int.scoreAlong((trees, step)) {
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var count = 0;
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for idx in trees.domain by step {
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count += 1;
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if trees[idx] >= this then break;
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}
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return count;
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}
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// Finally, we iterate (in parallel) over each tree, and tally up if it's
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// visible, as well as compute and note its score.
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var visible = 0;
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var bestScore = 0;
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forall coord in shapedNums.domain with (+ reduce visible, max reduce bestScore) {
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const tree = shapedNums[coord];
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visible += || reduce tree.visibleAlong(eachDirection(coord));
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bestScore reduce= * reduce tree.scoreAlong(eachDirection(coord));
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}
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writeln(visible);
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writeln(bestScore);
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