UVa 11468 Substring (AC自动机+概率DP)

时间：2022-12-16 00:15:31

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;

}

　　

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