UVA 11468 - Substring

AC自动机+概率

Given a set of pattern strings, and a text, you have to find, if any of the pattern is a substring of the text. If any of the pattern string can be found in text, then print “yes”, otherwise “no” (without quotes).

But, unfortunately, that’s not what is asked here. J

The problem described above, requires a input file generator. The generator generates a text of length L, by choosing L characters randomly. Probability of choosing each character is given as priori, and independent of choosing others.

Now, given a set of patterns, calculate the probability of a valid program generating “no”.

 

Input

First line contains an integer T, the number of test cases. Each case starts with an integer K, the number of pattern strings. Next K lines each contain a pattern string, followed by an integer N, number of valid characters.  Next N lines each contain a character and the probability of selecting that character, p_i. Next an integer L, the length of the string generated. The generated text can consist of only the valid characters, given above.

 

There will be a blank line after each test case.

 

Output

For each test case, output the number of test case, and the probability of getting a “no”.

 

Constraints

·         T  ≤ 50

·         K ≤ 20

·        Length of each pattern string is between 1 and 20

·        Each pattern string consists of only alphanumeric characters (‘a’ to ‘z’, ‘A’ to ‘Z’,’0’ to ‘9’)

·        Valid characters are all alphanumeric characters

·        ∑p_i = 1

·        L ≤ 100

Sample Input                             Output for Sample Input

2 1 a 2 a 0.5 b 0.5 2   2 ab ab 2 a 0.2 b 0.8 2

Case #1: 0.250000 Case #2: 0.840000

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define prt(k) cout<<#k"="<<k<<endl;
#define ll long long
#include<queue>
const int N=1e5+8;
const int SIZE=133;
#include<map>
int match[N];
double d[2222][111],vis[2222][333];
double prob[333];
int n,k;
map<char,int> mp;
    int ch[N][SIZE];
    int sz;
    void init()
    {
        sz=1;
        memset(ch[0],0,sizeof ch[0]);
        memset(match,0,sizeof match);
    }
    int f[N];
    void insert(char* s)
    {
        int u=0,n=strlen(s);//s.size();
        for(int i=0; i<n; i++)
        {
            int c= mp[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]=1;
    }
    int getfail()
    {
        queue<int> q;
        f[0]=0;
        for(int c=0; c<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<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]];
            }
        }
    }
double solve(int u,int L)
{
    if(L==0) return 1;
    if(vis[u][L]) return d[u][L];
    vis[u][L]=1;
    double &ans=d[u][L];
    ans=0;
    for(int i=0;i<n;i++)
        if(!match[ch[u][i]]) ans+=prob[i]*solve(ch[u][i],L-1);
    return ans;
}

int main()
{
    int _; cin>>_; int ca=1;
    while(_--)
    {
        cin>>k; mp.clear();
        char s[111][111];
        init();
        for(int i=1;i<=k;i++)
        {
            scanf("%s",s[i]);
        }
        cin>>n;
        for(int i=0;i<n;i++)
        {
            char t[7]; double a;
            cin>>t>>a;
            mp[t[0]]=i;
            prob[i]=a;
        }
        int L;
        init();
        for(int i=1;i<=k;i++)
            insert(s[i]);
        cin>>L;
        getfail();
        memset(vis,0,sizeof vis);
        double ans=solve(0,L);       
        printf("Case #%d: %.6lf\n",ca++,ans);
    }
}