UVa 11468 Substring (AC自动机+概率DP)

题意：给出一个字母表以及每个字母出现的概率。再给出一些模板串S。从字母表中每次随机拿出一个字母，一共拿L次组成一个产度为L的串，

问这个串不包含S中任何一个串的概率为多少？

析：先构造一个AC自动机，然后随机生成L个字母，就是在AC自动机的某个结点走多少步，dp[i][j] 表示在 i 结点，并且剩下 j 步，

然后记忆化搜索就OK了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
const int maxnode = 20 * 20 + 10;
const int sigma_size = 128;

struct Aho_Corasick{
  int ch[maxnode][sigma_size];
  int f[maxnode];
  bool match[maxnode];
  int sz;
  void init(){ sz = 1;  memset(ch[0], 0, sizeof ch[0]); }

  void insert(char *s){
    int u = 0;
    for(int i = 0; s[i]; ++i){
      int c = s[i];
      if(!ch[u][c]){
        memset(ch[sz], 0, sizeof ch[sz]);
        match[sz] = 0;
        ch[u][c] = sz++;
      }
      u = ch[u][c];
    }
    match[u] = true;
  }

  int getFail(){
    queue<int> q;
    f[0] = 0;
    for(int c = 0; c < sigma_size; ++c){
      int u = ch[0][c];
      if(u){  f[u] = 0;   q.push(u);  }
    }

    while(!q.empty()){
      int r = q.front();  q.pop();
      for(int c = 0; c < sigma_size; ++c){
        int u = ch[r][c];
        if(!u){ ch[r][c] = ch[f[r]][c];  continue; }
        q.push(u);
        int v = f[r];
        while(v && !ch[v][c])  v = f[v];
        f[u] = ch[v][c];
        match[u] |= match[f[u]];
      }
    }
  }

};
Aho_Corasick ac;
double dp[maxnode][110];
bool vis[maxnode][110];
char s[100];
struct Node{
  int x;  double p;
};
vector<Node> v;

double dfs(int u, int L){
  if(!L)  return 1.0;
  double &ans = dp[u][L];
  if(vis[u][L])  return ans;
  vis[u][L] = 1;
  ans = 0.0;
  for(int i = 0; i < n; ++i){
    int c = v[i].x;
    if(!ac.match[ac.ch[u][c]])  ans += dfs(ac.ch[u][c], L-1) * v[i].p;
  }
  return ans;
}

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);
    ac.init();
    for(int i = 0; i < n; ++i){
      scanf("%s", s);
      ac.insert(s);
    }
    ac.getFail();
    scanf("%d", &n);
    v.clear();
    for(int i = 0; i < n; ++i){
      double pp;
      scanf("%s %lf", s, &pp);
      v.push_back((Node){s[0], pp});
    }
    int L;
    scanf("%d", &L);
    memset(vis, 0, sizeof vis);
    printf("Case #%d: %f\n", kase, dfs(0, L));
  }
  return 0;
}