【POJ 2778】DNA Sequence

 1 #include <cstdio>
 2 #include <cstring>
 3 struct Matrix {
 4     int a[110][110];
 5 } mat, ti;
 6 int n, m, len, last, tn, ans, f[110], t[110][4], v[110];
 7 char str[11];
 8 void BFS ()
 9 {
10     int q[110], hd, tl;
11     for (q[hd = tl = 0] = 0; hd <= tl; hd++)
12         for (int i = 0; i < 4; i++)
13             t[q[hd]][i] ? (q[hd] ? f[t[q[hd]][i]] = t[f[q[hd]]][i] : 0) ,v[t[q[hd]][i]] |= v[f[t[q[hd]][i]]], q[++tl] = t[q[hd]][i] : t[q[hd]][i] = t[f[q[hd]]][i];
14 }
15 inline Matrix times (Matrix m1, Matrix m2)
16 {
17     Matrix ret;
18     memset (ret.a, 0, sizeof (ret.a));
19     for (int i = 0; i <= tn; i++)
20         for (int j = 0; j <= tn; j++)
21             for (int k = 0; k <= tn; ret.a[i][j] %= 100000, k++)
22                 ret.a[i][j] += (long long) m1.a[i][k] * m2.a[k][j] % 100000;
23     return ret;
24 }
25 inline int gc (char ch)
26 {
27     return ch == ‘A‘ ? 0 : ch == ‘C‘ ? 1 : ch == ‘T‘ ? 2 : 3;
28 }
29 int main ()
30 {
31     scanf ("%d %d", &n, &m);
32     for (int i = 0; i < n; i++)
33     {
34         scanf ("%s", (char*) &str);
35         len = strlen (str);
36         last = 0;
37         for (int j = 0, ch = gc (str[j]); j < len; ch = gc (str[++j]))
38             last = t[last][ch] ? t[last][ch] : t[last][ch] = ++tn;
39         v[last] = 1;
40     }
41     BFS ();
42     for (int i = 0; i <= tn; i++)
43         for (int j = 0; j < 4; j++)
44             (!v[t[i][j]]) && !v[i] ? mat.a[i][t[i][j]]++ : 0;
45     for (int i = 0; i <= tn; i++)    ti.a[i][i] = 1;
46     for (; m; m >>= 1)
47         (m & 1 ? ti = times (ti, mat) : ti), mat = times (mat, mat);
48     for (int i = 0; i <= tn; i++)    ans += ti.a[0][i];
49     printf ("%d", ans % 100000);
50 }