Substring UVA - 11468 AC自动机+概率DP

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 ;
 }