HDU 3689 Infinite monkey theorem [KMP DP]

Infinite monkey theorem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1702 Accepted Submission(s):
882

Problem Description

Could you imaging a monkey writing computer programs?
Surely monkeys are smart among animals. But their limited intelligence is no
match for our human beings. However, there is a theorem about monkeys, and it
states that monkeys can write everything if given enough time.
The theorem is
called “Infinite monkey theorem”. It states that a monkey hitting keys at random
on a typewriter keyboard for an infinite amount of time will almost surely type
any given text, which of course includes the programs you are about to write
(All computer programs can be represented as text, right?).
It’s very easy to
prove this theorem. A little calculation will show you that if the monkey types
for an infinite length of time the probability that the output contains a given
text will approach 100%.
However, the time used is too long to be physically
reasonable. The monkey will not be able to produce any useful programs even if
it types until the death of the universe. To verify this and ensure that our
human beings are not replaceable by monkeys, you are to calculate the
probability that a monkey will get things right.

Input

There will be several test cases.
Each test case
begins with a line containing two integers n and m separated by a whitespace
(2<=n<=26, 1<=m<=1000). n is the number of keys on the typewriter
and the monkey will hit these keys m times. Thus the typewriter will finally
produce an output of m characters.
The following n lines describe keys on the
typewriter. Each line has a lower case letter and a real number separated by a
whitespace. The letter indicates what the typewriter will produce if the monkey
hits that key and the real number indicates the probability that the monkey will
hit this key. Two hits of the monkey are independent of each other (Two
different hits have the same probability for a same key), and sum of all the
probabilities for each key is ensured to be 1.
The last line of the test case
contains a word composed of lower case letters. The length of the word will be
less than or equal to 10.
The input will end with a line of two zeros
separated by a whitespace. This line should not be processed.

Output

For each test case, output one line containing the
probability that the given word will appear in the typewriter’s output. The
output should be in percentage format and numbers should be rounded to two
digits after the decimal point.

Source

2010
Asia Hangzhou Regional Contest

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题意：

字符集中有一些字符（最多26个），给出每个字符的出现概率（它们的和保证为1）
再给出一个子串B
求：任给一个长度为N的字符串A（只能包含字符集中的字符），使得B是A的子串的概率。

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一边生成A，一边用KMP匹配B

f[i][j]表示生成到i位，当前匹配到B的j位的概率

枚举下一个字符，然后用KMP匹配就行了....匹配到now，就是f[i+1][now]+=f[i][j]*p[k]

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=,M=;
int cn,n,m,fail[N];
char c[],b[N],key[N];
double p[N],f[N][M];
void getFail(int n,char s[]){
    fail[]=;
    for(int i=;i<=n;i++){
        int j=fail[i-];
        while(j&&s[j+]!=s[i]) j=fail[j];
        fail[i]=s[j+]==s[i]?j+:;
    }
}
void dp(){
    memset(f,,sizeof(f));
    f[][]=;
    for(int i=;i<n;i++)
        for(int j=;j<m;j++) if(f[i][j])
            for(int k=;k<=cn;k++){
                int now=j;
                while(now&&b[now+]!=key[k]) now=fail[now];
                now+=b[now+]==key[k];
                f[i+][now]+=f[i][j]*p[k];
            }
    double ans=;
    for(int i=;i<=n;i++) ans+=f[i][m];
    ans*=;
    printf("%.2lf%%\n",ans);
}
int main(){
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d",&cn,&n)!=EOF&&(cn||n)){
        for(int i=;i<=cn;i++){
            scanf("%s%lf",c,&p[i]);
            key[i]=c[];
        }
        scanf("%s",b+);
        m=strlen(b+);
        getFail(m,b);
        dp();
    }
}