POJ 2778 DNA Sequence(AC自动机+矩阵快速幂)

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DNA Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13522 Accepted: 5152

Description

It's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence，For example, if a animal's DNA sequence contains segment ATC then it may mean that the animal may have a genetic disease. Until now scientists have found several those segments, the problem is how many kinds of DNA sequences of a species don't contain those segments. 

Suppose that DNA sequences of a species is a sequence that consist of A, C, T and G，and the length of sequences is a given integer n. 

Input

First line contains two integer m (0 <= m <= 10), n (1 <= n <=2000000000). Here, m is the number of genetic disease segment, and n is the length of sequences. 

Next m lines each line contain a DNA genetic disease segment, and length of these segments is not larger than 10. 

Output

An integer, the number of DNA sequences, mod 100000.

Sample Input

4 3ATACAGAA

Sample Output

36

Source

POJ Monthly--2006.03.26,dodo

题解: 我们可以通过AC自动机来寻找一个状态转移矩阵, ary[i][j]表示从第i个节点到第j个节点有多少种走法. 如何构造这个矩阵就是关键, 我们可以通过fail指针, 分别找出这个节点到A, C, T, G四个节点的走法, 从而构造出这个矩阵, 如代码所示. 另外需要注意的一点是, 如果fail指针指向的节点被感染了, 那么这个节点一定会被感染的.

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int N = 110;const int MOD = 100000;typedef long long LL;struct Matrix{int ary[N][N];int row, col;void init() {memset(ary, 0, sizeof(ary));}Matrix (int row = N, int col = N) {this->row = row, this->col = col;init();}Matrix operator=(const Matrix & A) {row = A.row;col = A.col;for (int i = 0; i < row; ++i)for (int j = 0; j < col; ++j)ary[i][j] = A.ary[i][j];}};const Matrix operator*(const Matrix & A, const Matrix & B) {Matrix t;t.row = A.row;t.col = B.col;for (int i = 0; i < A.row; ++i)for (int j = 0; j < B.col; ++j) {for (int k = 0; k < A.col; ++k)t.ary[i][j] += (int)(((LL)A.ary[i][k] * (LL)B.ary[k][j]) % MOD);t.ary[i][j] %= MOD;}return t;}struct Trie{int next[N][4], fail[N];bool vis[N];int total, root;Matrix mt;int new_node() {for (int i = 0; i < 4; ++i)next[total][i] = -1;vis[total] = false;return total++;}void init() {total = 0;mt.init();root = new_node();}int get_index(char ch) {if (ch == 'A') return 0;if (ch == 'T') return 1;if (ch == 'G') return 2;return 3;}void insert(char *str) {int cur = root;while (*str) {int idx = get_index(*str);if (next[cur][idx] == -1)next[cur][idx] = new_node();cur = next[cur][idx];++str;}vis[cur] = true;}void build() {queue<int> q;fail[root] = -1;q.push(root);while (!q.empty()) {int cur = q.front();q.pop();if (fail[cur] != -1 && vis[fail[cur]] == true)vis[cur] = true;for (int i = 0; i < 4; ++i) {if (next[cur][i] == -1) next[cur][i] = fail[cur] == -1 ? root : next[fail[cur]][i];else {fail[next[cur][i]] = fail[cur] == -1 ? root : next[fail[cur]][i];q.push(next[cur][i]);}}}mt.row = mt.col = total;for (int i = 0; i < total; ++i)for (int j = 0; j < 4; ++j)if (vis[next[i][j]] == false)++mt.ary[i][next[i][j]];}}tree;int quick_pow(Matrix & mt, int n) {Matrix ans, tmp = mt;ans.row = mt.row, ans.col = mt.col;for (int i = 0; i < mt.row; ++i)ans.ary[i][i] = 1;while (n) {if (n & 1)ans = ans * tmp;n >>= 1;tmp = tmp * tmp;}int res = 0;for (int i = 0; i < mt.row; ++i)res += ans.ary[0][i];return res % MOD;}int main() {int n, m;while (~scanf("%d%d", &n, &m)) {tree.init();char str[20];for (int i = 0; i < n; ++i) {scanf("%s", str);tree.insert(str);}tree.build();int ans = quick_pow(tree.mt, m);printf("%d\n", ans);}return 0;}