UVA - 11468 Substring

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Substring

Input: Standard Input

Output: Standard Output

 

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

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Problem Setter: Manzurur Rahman Khan

每产生一个字母，相当与在AC自动机中任意走了一步。。。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <cmath>using namespace std;const int maxn=1700;int idx(char c){    if(c>='0'&&c<='9')    {        return c-'0';    }    else if(c>='a'&&c<='z')    {        return c-'a'+10;    }    else if(c>='A'&&c<='Z')    {        return c-'A'+10+26;    }}int chd[maxn][65],match[maxn],f[maxn],sz;bool vis[maxn][120];void init(){    sz=1;    memset(chd[0],0,sizeof(chd[0]));    memset(match,0,sizeof(match));    memset(f,0,sizeof(f));}void insert(char *p){    int u=0;    for(;*p;p++)    {        int t=idx(*p);        if(!chd[u][t])        {            memset(chd[sz],0,sizeof(chd[sz]));            chd[u][t]=sz++;        }        u=chd[u][t];    }    match[u]=1;}int getFail(){    queue<int> q;    f[0]=0;    for(int c=0;c<62;c++)    {        int u=chd[0][c];        if(u)        {            f[u]=0;q.push(u);        }    }    while(!q.empty())    {        int r=q.front();q.pop();        for(int c=0;c<62;c++)        {            int u=chd[r][c];            if(!u)            {                chd[r][c]=chd[f[r]][c]; continue;            }            q.push(u);            int v=f[r];            while(v&&!chd[v][c]) v=f[v];            f[u]=chd[v][c];            match[u]|=match[f[u]];        }    }}int K,N,L;double prob[100],dp[maxn][120];double getProb(int u,int l){    if(l==0) return 1.;    if(vis[u][l]) return dp[u][l];    vis[u][l]=1;    double a=0.;    for(int i=0;i<62;i++)    {        if(!match[chd[u][i]])            a+=prob[i]*getProb(chd[u][i],l-1);    }    return dp[u][l]=a;}int main(){    int t,cas=1;    char word[30];    scanf("%d",&t);    while(t--)    {        init();        scanf("%d",&K);        for(int i=0;i<K;i++)        {            scanf("%s",word);            insert(word);        }        getFail();        memset(prob,0,sizeof(prob));        scanf("%d",&N);        for(int i=0;i<N;i++)        {            double p;            scanf("%s%lf",word,&p);            prob[idx(word[0])]=p;        }        memset(dp,0,sizeof(dp));        memset(vis,0,sizeof(vis));        scanf("%d",&L);        printf("Case #%d: %lf\n",cas++,getProb(0,L));    }    return 0;}