aoc

8b.c (1941B)

---

 /*
 I hate this stupid problem. There are a bunch of unwritten properties
 of the paths that make the problem very simple:
 
 - Different ghosts' paths do not end up in the same Z-node.
 - Each ghost has exactly one Z-node in its path.
 - Each ghost meets a Z-node exactly once before entering a loop.
 - If a ghost encounters a Z-node after X steps, it will encounter it exactly
   every X steps; preperiods just end up aligning nicely.
 
 In practice, everything just works out so that the lcm of the periods
 is the solution, which is very much not the case for a general input.
 This code computes the periods of the paths of the different ghosts. Then
 you can figure out the solution with a pocket calculator, or by hand.
 */
 
 #include <stdio.h>
 #include <string.h>
 
 #define N 1000
 
 typedef struct { char last; int next[2]; } node_t;
 
 int v[N*N];
 
 int ind(char s[3], char m[][3], int n) {
 	for (int j = 0; j < n; j++)
 		if (s[0] == m[j][0] && s[1] == m[j][1] && s[2] == m[j][2])
 			return j;
 	return -1;
 }
 
 int findperiod(int i, node_t *nodes, char *dir, int n, int k) {
 	memset(v, 0, sizeof(int) * N*N);
 	for (int j = i, d = 0, l = 1, state = -1; ; d = (d+1)%k) {
 		j = nodes[j].next[dir[d] == 'R'];
 		state = j*k+d;
 		if (v[state]) return l - v[state];
 		else v[state] = l++;
 	}
 	return -1;
 }
 
 int main() {
 	node_t nodes[N];
 	char *buf, dir[N], line[N], name[N][3], lstr[N][3], rstr[N][3];
 	int k, n;
 
 	k = strlen(fgets(dir, N, stdin)) - 1;
 	fgets(line, N, stdin);
 	for (n = 0; (buf = fgets(line, N, stdin)) != NULL; n++) {
 		memcpy(name[n], buf, 3);
 		nodes[n].last = name[n][2];
 		while (*buf != '(') buf++;
 		memcpy(lstr[n], buf+1, 3);
 		while (*buf != ' ') buf++;
 		memcpy(rstr[n], buf+1, 3);
 	}
 
 	for (int i = 0; i < n; i++) {
 		nodes[i].next[0] = ind(lstr[i], name, n);
 		nodes[i].next[1] = ind(rstr[i], name, n);
 	}
 
 	for (int i = 0; i < n; i++)
 		if (nodes[i].last == 'A')
 			printf("%d\n", findperiod(i, nodes, dir, n, k));
 
 	return 0;
 }