[POJ2778]DNA Sequence（AC自动机 + DP + 矩阵优化）

传送门

AC自动机加DP就不说了

注意到 m <= 10，所以模式串很少。

而 n 很大就需要 log 的算法，很容易想到矩阵。

但是该怎么构建？

还是矩阵 A(i,j) = ∑A(i,k) * A(k,j)，那么i到j的方案数就是j到k的方案数称k到j的方案数，那么直接矩阵快速幂即可

#include <queue>
#include <cstdio>
#include <cstring>
#define N 100001
#define p 100000
#define LL long long

int n, m, cnt, ans;
int next[N][4], fail[N], f[101][N];
bool val[N];
char s[N];
std::queue <int> q;

struct Matrix
{
	int n, m;
	LL a[141][141];
	Matrix()
	{
		n = m = 0;
		memset(a, 0, sizeof(a));
	}
}res;

inline int idx(char x)
{
	if(x == 'A') return 0;
	if(x == 'C') return 1;
	if(x == 'T') return 2;
	if(x == 'G') return 3;
}

inline void insert()
{
	int i, x, now = 0, len = strlen(s + 1);
	for(i = 1; i <= len; i++)
	{
		x = idx(s[i]);
		if(!next[now][x])
			next[now][x] = ++cnt;
		now = next[now][x];
	}
	val[now] = 1;
}

inline void make_fail()
{
	int i, now;
	for(i = 0; i < 4; i++)
		if(next[0][i])
			q.push(next[0][i]);
	while(!q.empty())
	{
		now = q.front();
		q.pop();
		for(i = 0; i < 4; i++)
		{
			if(!next[now][i])
			{
				next[now][i] = next[fail[now]][i];
				continue;
			}
			fail[next[now][i]] = next[fail[now]][i];
			val[next[now][i]] |= val[next[fail[now]][i]];
			q.push(next[now][i]);
		}
	}
}

inline Matrix operator * (Matrix x, Matrix y)
{
	int i, j, k;
	Matrix ret;
	ret.n = x.n;
	ret.m = y.m;
	for(i = 0; i <= x.n; i++)
		for(j = 0; j <= y.m; j++)
			for(k = 0; k <= y.n; k++)
				ret.a[i][j] = (ret.a[i][j] + x.a[i][k] * y.a[k][j]) % p;
	return ret;
}

inline Matrix operator ^ (Matrix x, int y)
{
	int i;
	Matrix ret;
	res.n = ret.m = cnt;
	for(i = 0; i <= cnt; i++) ret.a[i][i] = 1;
	for(; y; y >>= 1)
	{
		if(y & 1) ret = ret * x;
		x = x * x;
	}
	return ret;
}

int main()
{
	int i, j, k;
	scanf("%d %d", &n, &m);
	for(i = 1; i <= n; i++)
	{
		scanf("%s", s + 1);
		insert();
	}
	make_fail();
	res.n = res.m = cnt;
	for(i = 0; i <= cnt; i++)
	{
		if(val[i]) continue;
		for(j = 0; j < 4; j++)
			if(!val[next[i][j]])
				res.a[i][next[i][j]]++;
	}
	res = res ^ m;
	for(i = 0; i <= cnt; i++)
		ans = (ans + res.a[0][i]) % p;
	printf("%d\n", ans);
	return 0;
}