【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。

 #include <cstdio>
 #include <cstring>

 #define MAXN 85

 typedef struct {
     char m[MAXN][MAXN];
 } mat_st;

 int n, m;
 char buf[MAXN];
 mat_st e;

 mat_st mat_mult(mat_st a, mat_st b) {
     int i, j, k;
     mat_st c;
     memset(c.m, , sizeof(c.m));

     for (i=; i<=n; ++i) {
         for (j=; j<=n; ++j) {
             for (k=; k<=n; ++k)
                 c.m[i][j] += a.m[i][k] & b.m[k][j];
         }
     }
     return c;
 }

 mat_st mat_power(mat_st a, int r) {
     mat_st ret;

     memset(ret.m, , sizeof(ret.m));
     for (int i=; i<=n; ++i)
         ret.m[i][i] = ;

     while (r) {
         if (r & )
             ret = mat_mult(ret, a);
         a = mat_mult(a, a);
         r >>= ;
     }
     return ret;
 }

 int main() {
     mat_st org;
     int i, j;

     while (scanf("%d %d",&n,&m)!=EOF && n) {
         memset(org.m, , sizeof(org.m));
         for (i=; i<=n; ++i) {
             scanf("%d", &j);
             org.m[i][j] = ;
         }
         getchar();
         gets(buf+);
         mat_st ans = mat_power(org, m);
         for (i=; i<=n; ++i) {
             for (j=; j<=n; ++j) {
                 if (ans.m[j][i]) {
                     printf("%c", buf[j]);
                     break;
                 }
             }
         }
         printf("\n");
     }

     return ;
 }