Day 20: Race Condition
Megathread guidelines
- Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
- You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL
FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
You are viewing a single thread.
View all comments 2 points
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
