poj 2778 DNA Sequence

题目传送门

题意：

给出m个DNA序列，求出长度为n的不含上述序列的个数。(0 <= m <= 10)，(1 <= n <=2000000000)

题解：

首先我们应将DNA序列存入Tire中，之后可以构造一个矩阵A,A[i][j]表示从i节点状态是否可以一步转移到j节点状态。构造方法则很明显：沿着fail指针进行构造，如果它是被标记的序列那么就置为0。所以在BuildFail是要将没有儿子节点通过fail指针指向前面的节点。之后求出A^n求出方案数，ans=sigma（A[0][i]）。

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

inline char idx(const char &c) {
switch (c) {
case 'A' : return 0;
case 'T' : return 1;
case 'C' : return 2;
case 'G' : return 3;
}
}

//AC-auto
const int MS = 110, Siz = 4;
int ch[MS][Siz], root, totN, Fail[MS];
bool Unable[MS];

int NewNode() {
Unable[++totN] = 0;
memset(ch[totN], 0, sizeof ch[totN]);
return totN;
}

void Insert(char *S) {
int cur = root;
char id;
while (*S) {
id = idx(*S);
if (ch[cur][id]) {
cur = ch[cur][id];
} else {
cur = ch[cur][id] = NewNode();
}
++S;
}
Unable[cur] = 1;
}

void BuildFail() {
queue<int>q;
Fail[root] = root;
for (int i = 0; i < Siz; ++i) if (ch[root][i]) {
Fail[ch[root][i]] = root;
q.push(ch[root][i]);
}
int f, p;
while (!q.empty()) {
p = q.front(); q.pop();
if (Unable[Fail[p]]) Unable[p] = 1;
for (int i = 0; i < Siz; ++i)
if (ch[p][i]) {
f = Fail[p];

while (!ch[f][i] && f) f = Fail[f];

if (ch[f][i]) Fail[ch[p][i]] = ch[f][i];
else Fail[ch[p][i]] = root;

q.push(ch[p][i]);
} else {
ch[p][i] = ch[Fail[p]][i];//构建假儿子
}

}

}

//Matrix

typedef long long LL;
const int Mod = 100000;

LL tmp1[MS][MS], tmp2[MS][MS];

void Matmul(LL A[MS][MS], LL B[MS][MS]){
memset(tmp1, 0, sizeof tmp1);
for (int i = 0; i <= totN; ++i)
for (int j = 0; j <= totN; ++j)
for (int k = 0; k <= totN; ++k)
tmp1[i][j] = (tmp1[i][j] + A[i][k] * B[k][j]) % Mod;;

memcpy(A, tmp1, sizeof tmp1);
}

void MatPow(LL A[MS][MS], int n) {
for (int i = totN; ~i; --i)
for (int j = totN; ~j; --j)
tmp2[i][j] = (i == j);
while (n) {
if (n & 1) Matmul(tmp2, A);
Matmul(A, A);
n >>= 1;
}
memcpy(A, tmp2, sizeof tmp2);
}

//

void BuildMat(LL A[MS][MS]) {
for (int i = 0; i <= totN; ++i)
for (int j = 0; j < Siz; ++j)
if (!Unable[i] && !Unable[ch[i][j]])//不是被标记的点才可以连接
++A[i][ch[i][j]];
}

char tmpS[12];

int main() {
int n, m, tot;
LL ans[MS][MS];
while (scanf("%d%d\n", &m, &n) == 2) {
totN = -1; root = NewNode();
while (m--) {
gets(tmpS);
Insert(tmpS);
}
BuildFail();
memset(ans, 0, sizeof ans);
BuildMat(ans);

for (int i = 0; i <= totN; ++i) {
for (int j = 0; j <= totN; ++j)
printf("%d ", ans[i][j]);
puts("");
}
tot = 0;
MatPow(ans, n);
for (int i = 0; i <= totN; ++i) {
tot += ans[0][i];
tot %= Mod;
}
printf("%d\n", tot);
}
return 0;
}