POJ 2778 DNA Sequence（AC自动机+矩阵快速幂）

题目链接：http://poj.org/problem?id=2778

题意：有m种DNA序列是有疾病的，问有多少种长度为n的DNA序列不包含任何一种有疾病的DNA序列。（仅含A,T,C,G四个字符）

思路：Trie图的状态转移，用矩阵mat[i][j]来表示从结点i到j只走一步有几种走法，那么mat的n次幂就表示从结点i到j走n步有几种走法，题目要求解的就是从头节点走n步且不包含危险结点的走法。

mat = mat^n   ans = (mat[0][0] + mat[0][1] + ... + mat[0][num]) num为结点个数

code：

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <map>
 using namespace std;
 const int KIND = ;
 const int MAXN = ;
 const int MOD = ;
 typedef long long LL;
 
 struct Trie
 {
     int next[MAXN][KIND], fail[MAXN];
     bool isExit[MAXN];
     int root, L;
     map<char, int> mp;
     LL mat[MAXN][MAXN];
     LL ret[MAXN][MAXN];
     LL tmp[MAXN][MAXN];
     int create()
     {
         for (int i = ; i < KIND; ++i)
             next[L][i] = -;
         isExit[L++] = false;
         return L - ;
     }
     void init()
     {
         L = ;
         root = create();
         mp['A'] = ;
         mp['C'] = ;
         mp['G'] = ;
         mp['T'] = ;
         memset(mat, , sizeof(mat));
         memset(ret, , sizeof(ret));
     }
     void insert(char str[])
     {
         int now = root;
         int len = strlen(str);
         for (int i = ; i < len; ++i)
         {
             if (- == next[now][mp[str[i]]])
                 next[now][mp[str[i]]] = create();
             now = next[now][mp[str[i]]];
         }
         isExit[now] = true;
     }
     void build()
     {
         queue<int>Q;
         for (int i = ; i < KIND; ++i)
         {
             if (- == next[root][i])
                 next[root][i] = root;
             else
             {
                 fail[next[root][i]] = root;
                 Q.push(next[root][i]);
             }
         }
         while (!Q.empty())
         {
             int now = Q.front();
             Q.pop();
             if (isExit[fail[now]])
                 isExit[now] = true;
             for (int i = ; i < KIND; ++i)
             {
                 if (- == next[now][i])
                     next[now][i] = next[fail[now]][i];
                 else
                 {
                     fail[next[now][i]] = next[fail[now]][i];
                     Q.push(next[now][i]);
                 }
             }
         }
     }
     void getMatrix()
     {
         for (int i = ; i < L; ++i)
         {
             for (int j = ; j < KIND; ++j)
             {
                 if (!isExit[next[i][j]])
                     ++mat[i][next[i][j]];
             }
         }
     }
     void matrixMul(LL mat1[MAXN][MAXN], LL mat2[MAXN][MAXN])
     {
         LL mat3[MAXN][MAXN];
         for (int i = ; i < L; ++i)
         {
             for (int j = ; j < L; ++j)
             {
                 mat3[i][j] = ;
                 for (int k = ; k < L; ++k)
                     mat3[i][j] = (mat3[i][j] + mat1[i][k] * mat2[k][j]) % MOD;
             }
         }
         memcpy(mat1, mat3, sizeof(mat3));
     }
     void matrixQuickMod(LL n)
     {
         getMatrix();
         for (int i = ; i < L; ++i)
         {
             ret[i][i] = ;
             for (int j = ; j < L; ++j)
                 tmp[i][j] = mat[i][j];
         }
         while (n)
         {
             if (n & ) matrixMul(ret, tmp);
             matrixMul(tmp, tmp);
             n >>= ;
         }
     }
 };
 Trie ac;
 char str[];
 int main()
 {
     int m;
     LL n;
     while (scanf("%d %lld", &m, &n) != EOF)
     {
         ac.init();
         for (int i = ; i < m; ++i)
         {
             scanf("%s", str);
             ac.insert(str);
         }
         ac.build();
         ac.matrixQuickMod(n);
         int ans = ;
         for (int i = ; i < ac.L; ++i)
             ans = (ans + ac.ret[][i]) % MOD;
         printf("%d\n", ans);
     }
     return ;
 }