Advent Of Code

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Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

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🏃‍♀️ - 2024 DAY 20 SOLUTIONS -🏃‍♀️ (self.advent_of_code)

submitted 2 days ago by CameronDev to c/advent_of_code

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Day 20: Race Condition

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[–] [email protected] 2 points 1 day ago

C

Note to self: reread the puzzle after waking up properly! I initially wrote a great solution for the wrong question (pathfinding with a given number of allowed jumps).

For the actual question, a bit boring but flood fill plus some array iteration saved the day again. Find the cost for every open tile with flood fill, then for each position and offset combination, see if that jump yields a lower cost at the destination.

For arbitrary inputs this would require eliminating the non-optimal paths, but the one path covered all open tiles.

Code

#include "common.h"

#define GB 2	/* safety border on X/Y plane */
#define GZ (GB+141+2+GB)

static char G[GZ][GZ];		/* grid */
static int  C[GZ][GZ];		/* costs */
static int sx,sy, ex,ey;	/* start, end pos */

static void
flood(int x, int y)
{
	int lo = INT_MAX;

	if (x<1 || x>=GZ-1 ||
	    y<1 || y>=GZ-1 || G[y][x]!='.')
		return;

	if (C[y-1][x]) lo = MIN(lo, C[y-1][x]+1);
	if (C[y+1][x]) lo = MIN(lo, C[y+1][x]+1);
	if (C[y][x-1]) lo = MIN(lo, C[y][x-1]+1);
	if (C[y][x+1]) lo = MIN(lo, C[y][x+1]+1);

	if (lo != INT_MAX && (!C[y][x] || lo < C[y][x])) {
		C[y][x] = lo;

		flood(x, y-1); flood(x-1, y);
		flood(x, y+1); flood(x+1, y);
	}
}

static int
count_shortcuts(int lim, int min)
{
	int acc=0, x,y, dx,dy;

	for (y=GB; y<GZ-GB; y++)
	for (x=GB; x<GZ-GB; x++)
	for (dy = -lim; dy <= lim; dy++)
	for (dx = abs(dy)-lim; dx <= lim-abs(dy); dx++)
		acc += x+dx >= 0 && x+dx < GZ &&
		       y+dy >= 0 && y+dy < GZ && C[y][x] &&
		       C[y][x]+abs(dx)+abs(dy) <= C[y+dy][x+dx]-min;

	return acc;
}

int
main(int argc, char **argv)
{
	int x,y;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));
	for (y=2; fgets(G[y]+GB, GZ-GB*2, stdin); y++)
		assert(y < GZ-3);

	for (y=GB; y<GZ-GB; y++)
	for (x=GB; x<GZ-GB; x++)
		if (G[y][x] == 'S') { G[y][x]='.'; sx=x; sy=y; } else
		if (G[y][x] == 'E') { G[y][x]='.'; ex=x; ey=y; }

	C[sy][sx] = 1;

	flood(sx, sy-1); flood(sx-1, sy);
	flood(sx, sy+1); flood(sx+1, sy);

	printf("20: %d %d\n",
	    count_shortcuts(2, 100),
	    count_shortcuts(20, 100));

	return 0;
}

https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day20.c