Substring

Substring

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

2

1

a

2

a 0.5

b 0.5

2

2

ab

ab

2

a 0.2

b 0.8

2

分析：ac自动机+dp;

　　　dp[i][j]表示到i节点剩余j长度的概率，则dp[i][j]+=p[j]*dp[ch[i][k]][j-1];

　　　记忆化搜索优化;

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
#define intxt freopen("in.txt","r",stdin)
const int maxn=1e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,cas,vis[maxn][105];
char a[maxn];
double p[maxn],q[maxn][105];
struct node
{
    int ch[maxn][65],f[maxn],val[maxn],last[maxn],cnt[maxn],sz;
    void init()
    {
        sz=0;
        memset(ch[0],0,sizeof(ch[0]));
        memset(cnt,0,sizeof(cnt));
    }
    //字符编号;
    int idx(char x)
    {
        if(x>='0'&&x<='9')return x-'0';
        else if(x>='A'&&x<='Z')return x-'A'+10;
        else if(x>='a'&&x<='z')return x-'a'+36;
    }
    //插入字符串,v非0;
    void insert(char *p,int v)
    {
        int pos=0;
        for(int i=0;p[i];i++)
        {
            int x=idx(p[i]);
            if(!ch[pos][x])
            {
                ch[pos][x]=++sz;
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz]=0;
            }
            pos=ch[pos][x];
        }
        val[pos]=1;
    }
    //递归打印以结点x结尾的所有字符串;
    void out(int x)
    {
        if(x)
        {
            cnt[val[x]]++;
            out(last[x]);
        }
    }
    //找模板;
    void find(char *p)
    {
        int pos=0;
        for(int i=0;p[i];i++)
        {
            int x=idx(p[i]);
            pos=ch[pos][x];
            if(val[pos])out(pos);
            else if(last[pos])out(last[pos]);
        }
    }
    //fail函数;
    void fail()
    {
        queue<int>p;
        f[0]=0;
        //初始化;
        for(int i=0;i<65;i++)
        {
            int x=ch[0][i];
            if(x)f[x]=0,p.push(x),last[x]=0;
        }
        //bfs计算fail;
        while(!p.empty())
        {
            int x=p.front();
            p.pop();
            for(int i=0;i<65;i++)
            {
                int y=ch[x][i];
                if(!y)
                {
                    ch[x][i]=ch[f[x]][i];
                    continue;
                }
                p.push(y);
                int v=f[x];
                while(v&&!ch[v][i])v=f[v];
                f[y]=ch[v][i];
                last[y]=val[f[y]]?f[y]:last[f[y]];
                val[y]|=val[f[y]];
            }
        }
    }
}ac;
double gao(int x,int y)
{
    if(!y)return 1.0;
    if(vis[x][y])return q[x][y];
    vis[x][y]=1;
    double &ans=q[x][y];
    ans=0.0;
    for(int i=0;i<65;i++)
    {
        if(!ac.val[ac.ch[x][i]])ans+=p[i]*gao(ac.ch[x][i],y-1);
    }
    return ans;
}
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        ac.init();
        rep(i,0,64)p[i]=0;
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        rep(i,1,n)
        {
            scanf("%s",a);
            ac.insert(a,i);
        }
        ac.fail();
        scanf("%d",&n);
        rep(i,1,n)
        {
            double x;
            scanf("%s%lf",a,&x);
            p[ac.idx(a[0])]=x;
        }
        scanf("%d",&m);
        printf("Case #%d: %.6f\n",++cas,gao(0,m));
    }
    //system("Pause");
    return 0;
}