题意：
给出一些字符和各自对应的选择概率，随机选择L次后得到一个长度为L的随机字符串S。
给出K个模板串，计算S不包含任何一个模板串的概率
dp【i】【j】表示走到AC自动机 i 这个节点 还需要走 j 步的概率。
表示不会概率DP ，看网上题解写的。
通过记忆化搜索去写。
注意一点字符有大小字母和数字
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map> #define pi acos(-1.0)
#define eps 1e-9
#define fi first
#define se second
#define rtl rt<<1
#define rtr rt<<1|1
#define bug printf("******\n")
#define mem(a, b) memset(a,b,sizeof(a))
#define name2str(x) #x
#define fuck(x) cout<<#x" = "<<x<<endl
#define sfi(a) scanf("%d", &a)
#define sffi(a, b) scanf("%d %d", &a, &b)
#define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c)
#define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define sfL(a) scanf("%lld", &a)
#define sffL(a, b) scanf("%lld %lld", &a, &b)
#define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c)
#define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
#define sfs(a) scanf("%s", a)
#define sffs(a, b) scanf("%s %s", a, b)
#define sfffs(a, b, c) scanf("%s %s %s", a, b, c)
#define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
#define FIN freopen("../in.txt","r",stdin)
#define gcd(a, b) __gcd(a,b)
#define lowbit(x) x&-x
#define IO iOS::sync_with_stdio(false) using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const ULL seed = ;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + ;
const int maxm = 8e6 + ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int T, n, m, L, id[], vis[][];
char buf[];
double dp[][], pro[]; int get_num(char ch) {
if (ch >= 'a' && ch <= 'z') return ch - 'a';
if (ch >= 'A' && ch <= 'Z') return ch - 'A' + ;
if (ch >= '' && ch <= '') return ch - '' + ;
} struct Aho_Corasick {
int next[][], fail[], End[];
int root, cnt; int newnode() {
for (int i = ; i < ; i++) next[cnt][i] = -;
End[cnt++] = ;
return cnt - ;
} void init() {
cnt = ;
root = newnode();
} void insert(char buf[]) {
int len = strlen(buf);
int now = root;
for (int i = ; i < len; i++) {
if (next[now][get_num(buf[i])] == -) next[now][get_num(buf[i])] = newnode();
now = next[now][get_num(buf[i])];
}
End[now] = ;
} void build() {
queue<int> Q;
fail[root] = root;
for (int i = ; i < ; 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();
End[now] |= End[fail[now]];
for (int i = ; i < ; 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]);
}
}
} double solve(int pos, int res) {
if (!res) return 1.0;
if (vis[pos][res]) return dp[pos][res];
vis[pos][res] = ;
double &ret = dp[pos][res];
ret = ;
for (int i = ; i < m; ++i) {
int idx = id[i];
if (!End[next[pos][idx]]) ret += pro[i] * solve(next[pos][idx], res - );
}
return ret;
} void debug() {
for (int i = ; i < cnt; i++) {
printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
for (int j = ; j < ; j++) printf("%2d", next[i][j]);
printf("]\n");
}
}
} ac; int main() {
// FIN;
int cas = ;
sfi(T);
while (T--) {
sfi(n);
ac.init();
for (int i = ; i < n; ++i) {
sfs(buf);
ac.insert(buf);
}
ac.build();
sfi(m);
for (int i = ; i < m; ++i) {
scanf("%s%lf", buf, &pro[i]);
id[i] = get_num(buf[]);
}
sfi(L);
mem(vis, );
printf("Case #%d: %.6f\n", cas++, ac.solve(, L));
}
return ;
}