1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
|
/*
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 <string.h>
#define N 200
#define S 202300L
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;
}
|
