题目地址
因为n很大，很自然想到用矩阵快速幂。
首先要知道对于一个01矩阵，如果m[i, j] = 1,表示从i到j有一条路，那么这个矩阵在自乘n次后，m[i,j]表示从i->j走n步的的方案数（离散书上的图论相关内容）
所以只需要找到哪些边是可以走的，然后跑一下矩阵快速幂就行了。
先把病毒放到ac自动机里面，对于结尾的节点标记一下，然后把可以走的边放到矩阵里面跑一下矩阵快速幂就行了。
#include<queue>
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int mod = 100000;
typedef long long ll;
int Hash[125];
struct AC{
int nex[300][4], tot, root;
int f[300], ed[300];
int newnode() {
for(int i = 0; i < 4; i++) nex[tot][i] = -1;
ed[tot] = 0;
return tot++;
}
void init() {
tot = 0;
root = newnode();
}
void insert(char *s) {
int u = root, len = strlen(s);
for(int i = 0; i < len; i++) {
int ch = Hash[s[i]];
if(nex[u][ch] == -1) nex[u][ch] = newnode();
u = nex[u][ch];
}
ed[u] = 1;
}
void getfail() {
queue<int>Q;
for(int i = 0; i < 4; i++) {
if(nex[root][i] == -1) nex[root][i] = root;
else {
f[nex[root][i]] = root;
Q.push(nex[root][i]);
}
}
while(!Q.empty()) {
int u = Q.front();Q.pop();
ed[u] |= ed[f[u]];
for(int i = 0; i < 4; i++) {
if(nex[u][i] == -1) nex[u][i] = nex[f[u]][i];
else {
f[nex[u][i]] = nex[f[u]][i];
Q.push(nex[u][i]);
}
}
}
}
}ac;
class matrix{
public:
ll a[105][105];
int n, m;
matrix(int n, int m) {
this->n = n;
this->m = m;
memset(a, 0, sizeof(a));
}
matrix operator *(matrix &b) {
matrix c(n, b.m);
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= b.m; j++) {
for(int k = 1; k <= m; k++) {
c.a[i][j] += a[i][k] * b.a[k][j];
if (c.a[i][j] > mod) {
c.a[i][j] %= mod;
}
}
}
}
return c;
}
matrix pow(int x) {
matrix res(n, n), A(n, n);
for(int i = 1; i <= n; i++) res.a[i][i] = 1;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
A.a[i][j] = a[i][j];
}
}
while(x) {
if(x&1) res = res*A;
A = A*A;
x >>= 1;
}
return res;
}
};
char ss[10];
int m, n;
int main() {
Hash['A'] = 0;Hash['C'] = 1;Hash['T'] = 2;Hash['G'] = 3;
while(scanf("%d%d", &m, &n) == 2) {
ac.init();
for(int i = 1; i <= m; i++) {
scanf("%s", ss);
ac.insert(ss);
}
ac.getfail();
matrix A(ac.tot, ac.tot);
for(int i = 0; i < ac.tot; i++) {
if(ac.ed[i]) continue;
for(int j = 0; j < 4; j++) {
int t = ac.nex[i][j];
if(!ac.ed[t]) A.a[i+1][t+1]++;
}
}
matrix ans(ac.tot, ac.tot);
ans = A.pow(n);
ll sum = 0;
for (int i = 1; i <= ac.tot; ++i) {
sum = sum + ans.a[1][i];
if (sum > mod) {
sum %= mod;
}
}
printf("%lld\n", sum);
}
}