From 709005f9a565405d9413eebf8c4ccaff53f0654a Mon Sep 17 00:00:00 2001
From: Sebastiano Tronto <sebastiano@tronto.net>
Date: Fri, 22 Dec 2023 16:44:48 +0100
Subject: Added solutions for 21 and 22

---
 2023/21/21b.c | 83 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 83 insertions(+)
 create mode 100644 2023/21/21b.c

(limited to '2023/21/21b.c')

diff --git a/2023/21/21b.c b/2023/21/21b.c
new file mode 100644
index 0000000..1748185
--- /dev/null
+++ b/2023/21/21b.c
@@ -0,0 +1,83 @@
+/*
+This is another problem I did not enjoy. Once again, it was only solvable
+because of the carefully crafted input case. This makes the problem
+solvable, but at the same time it leaves you wondering - which unsaid
+assumptions are true and which are not?
+
+For example, in my first 14 attempts or so I assumed that every non-# cell
+was reached at the same time as if there were no #-cells. A reasonable
+assumption, and almost true. Unfortunately there were some cells that
+were completely enclosed by #-cells, and thus unreachable. See fuckyou.png
+and fuckyou2.png.
+*/
+
+#include <inttypes.h>
+#include <stdbool.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+#define N 200
+#define S 202300L /* My input was 202300*131+65 */
+
+char map[N][N];
+int64_t n;
+
+int go(char m[][N], int i, int j) {
+	if (i >= 0 && j >= 0 && i < n && j < n && m[i][j] == '.') {
+		m[i][j] = 'S';
+		return 1;
+	}
+	return 0;
+}
+
+int64_t flood(int64_t x, int64_t y, int64_t steps) {
+	int64_t c, i, j;
+	char tmp[N][N];
+	for (int i = 0; i < n; i++) memcpy(tmp[i], map[i], n*sizeof(char));
+	tmp[x][y] = 'S';
+	for (int s = 0; s < steps; s++) {
+		for (i = 0, c = 0; i < n; i++) {
+			for (j = (s+x+y+i)%2; j < n; j += 2) {
+				if (tmp[i][j] == 'S') {
+					tmp[i][j] = '.';
+					c += go(tmp, i-1, j) +
+					     go(tmp, i+1, j) +
+					     go(tmp, i, j-1) +
+					     go(tmp, i, j+1);
+				}
+			}
+		}
+	}
+	return c;
+}
+
+int main() {
+	for (n = 0; fgets(map[n], N, stdin) != NULL; n++) ;
+	map[n/2][n/2] = '.';
+
+	int64_t fulleven = flood(n/2, n/2, n-1);
+	int64_t fullodd = flood(n/2, n/2, n);
+	int64_t arrowr = flood(n/2, 0, n-1);
+	int64_t arrowl = flood(n/2, n-1, n-1);
+	int64_t arrowd = flood(0, n/2, n-1);
+	int64_t arrowu = flood(n-1, n/2, n-1);
+	int64_t topl = flood(0, 0, n/2-1);
+	int64_t topr = flood(0, n-1, n/2-1);
+	int64_t botl = flood(n-1, 0, n/2-1);
+	int64_t botr = flood(n-1, n-1, n/2-1);
+	int64_t cutbr = flood(0, 0, n/2+n-1);
+	int64_t cutbl = flood(0, n-1, n/2+n-1);
+	int64_t cuttr = flood(n-1, 0, n/2+n-1);
+	int64_t cuttl = flood(n-1, n-1, n/2+n-1);
+
+	int64_t fulls = S*S*fulleven + (S-1)*(S-1)*fullodd;
+	int64_t smallborder = S*(topl + topr + botl + botr);
+	int64_t bigborder = (S-1)*(cutbr + cutbl + cuttr + cuttl);
+	int64_t arrows = arrowr + arrowl + arrowu + arrowd;
+	int64_t final = fulls + smallborder + bigborder + arrows;
+
+	printf("%" PRId64 "\n", final);
+
+	return 0;
+}
-- 
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