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);    }}