【HDOJ】2371 Decode the Strings

快速矩阵乘法。注意，原始字符串即为decode后的字符串。题目是要找到原始串。

 1 #include <cstdio>
 2 #include <cstring>
 3
 4 #define MAXN 85
 5
 6 typedef struct {
 7     char m[MAXN][MAXN];
 8 } mat_st;
 9
10 int n, m;
11 char buf[MAXN];
12 mat_st e;
13
14 mat_st mat_mult(mat_st a, mat_st b) {
15     int i, j, k;
16     mat_st c;
17     memset(c.m, 0, sizeof(c.m));
18
19     for (i=1; i<=n; ++i) {
20         for (j=1; j<=n; ++j) {
21             for (k=1; k<=n; ++k)
22                 c.m[i][j] += a.m[i][k] & b.m[k][j];
23         }
24     }
25     return c;
26 }
27
28 mat_st mat_power(mat_st a, int r) {
29     mat_st ret;
30
31     memset(ret.m, 0, sizeof(ret.m));
32     for (int i=1; i<=n; ++i)
33         ret.m[i][i] = 1;
34
35     while (r) {
36         if (r & 1)
37             ret = mat_mult(ret, a);
38         a = mat_mult(a, a);
39         r >>= 1;
40     }
41     return ret;
42 }
43
44 int main() {
45     mat_st org;
46     int i, j;
47
48     while (scanf("%d %d",&n,&m)!=EOF && n) {
49         memset(org.m, 0, sizeof(org.m));
50         for (i=1; i<=n; ++i) {
51             scanf("%d", &j);
52             org.m[i][j] = 1;
53         }
54         getchar();
55         gets(buf+1);
56         mat_st ans = mat_power(org, m);
57         for (i=1; i<=n; ++i) {
58             for (j=1; j<=n; ++j) {
59                 if (ans.m[j][i]) {
60                     printf("%c", buf[j]);
61                     break;
62                 }
63             }
64         }
65         printf("\n");
66     }
67
68     return 0;
69 }

【HDOJ】2371 Decode the Strings