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1/*
2This is another problem I did not enjoy. Once again, it was only solvable
3because of the carefully crafted input case. This makes the problem
4solvable, but at the same time it leaves you wondering - which unsaid
5assumptions are true and which are not?
6
7For example, in my first 14 attempts or so I assumed that every non-# cell
8was reached at the same time as if there were no #-cells. A reasonable
9assumption, and almost true. Unfortunately there were some cells that
10were completely enclosed by #-cells, and thus unreachable. See fuckyou.png
11and fuckyou2.png.
12*/
13
14#include <inttypes.h>
15#include <stdbool.h>
16#include <stdio.h>
17#include <stdlib.h>
18#include <string.h>
19
20#define N 200
21#define S 202300L /* My input was 202300*131+65 */
22
23char map[N][N];
24int64_t n;
25
26int go(char m[][N], int i, int j) {
27 if (i >= 0 && j >= 0 && i < n && j < n && m[i][j] == '.') {
28 m[i][j] = 'S';
29 return 1;
30 }
31 return 0;
32}
33
34int64_t flood(int64_t x, int64_t y, int64_t steps) {
35 int64_t c, i, j;
36 char tmp[N][N];
37 for (int i = 0; i < n; i++) memcpy(tmp[i], map[i], n*sizeof(char));
38 tmp[x][y] = 'S';
39 for (int s = 0; s < steps; s++) {
40 for (i = 0, c = 0; i < n; i++) {
41 for (j = (s+x+y+i)%2; j < n; j += 2) {
42 if (tmp[i][j] == 'S') {
43 tmp[i][j] = '.';
44 c += go(tmp, i-1, j) +
45 go(tmp, i+1, j) +
46 go(tmp, i, j-1) +
47 go(tmp, i, j+1);
48 }
49 }
50 }
51 }
52 return c;
53}
54
55int main() {
56 for (n = 0; fgets(map[n], N, stdin) != NULL; n++) ;
57 map[n/2][n/2] = '.';
58
59 int64_t fulleven = flood(n/2, n/2, n-1);
60 int64_t fullodd = flood(n/2, n/2, n);
61 int64_t arrowr = flood(n/2, 0, n-1);
62 int64_t arrowl = flood(n/2, n-1, n-1);
63 int64_t arrowd = flood(0, n/2, n-1);
64 int64_t arrowu = flood(n-1, n/2, n-1);
65 int64_t topl = flood(0, 0, n/2-1);
66 int64_t topr = flood(0, n-1, n/2-1);
67 int64_t botl = flood(n-1, 0, n/2-1);
68 int64_t botr = flood(n-1, n-1, n/2-1);
69 int64_t cutbr = flood(0, 0, n/2+n-1);
70 int64_t cutbl = flood(0, n-1, n/2+n-1);
71 int64_t cuttr = flood(n-1, 0, n/2+n-1);
72 int64_t cuttl = flood(n-1, n-1, n/2+n-1);
73
74 int64_t fulls = S*S*fulleven + (S-1)*(S-1)*fullodd;
75 int64_t smallborder = S*(topl + topr + botl + botr);
76 int64_t bigborder = (S-1)*(cutbr + cutbl + cuttr + cuttl);
77 int64_t arrows = arrowr + arrowl + arrowu + arrowd;
78 int64_t final = fulls + smallborder + bigborder + arrows;
79
80 printf("%" PRId64 "\n", final);
81
82 return 0;
83}

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