Crazy Typewriter
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 391 Accepted Submission(s): 109
Problem Description
There was a crazy typewriter before. When the writer is not very sober, it would type characters randomly, one per second, the possiblities of characters may differ.
The protagonist in this problem wants to tell acmers some secrets, of course, by the typewriter.
There had been several opportunities, but the protagonist let them sliped. Now, another opportunity came, the writer started a new paragraph. The protagonist found
that he could set the possiblities of each character in happy astonishment. After the possiblities had been set, he wanted to know the exception of time at least the writer need to be mind-absent if any secret was typed out.
fewovigwnierfbiwfioeifaorfwarobahbgssjqmdowj
Input
There are several cases, no more than 15.
The first line of each case contains an integer n, no more than 15, indicating the number of secrets.
The second line contains 26 real numbers, indicating the set possibilities of ‘a‘-‘z‘, respectively, the sum would be 1.0 .
Then n lines, each contains a secret, no longer than 15, which is made up by lowercase letters ‘a‘-‘z‘.
Output
A single line contains the expectation of time for each case， in seconds with six decimal, if the exception doesn‘t exist, output "Infinity"
Sample Input
2
0.5000 0.5000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
ab
aa
Sample Output
3.000000
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <queue>
5 #include <algorithm>
6 #include <cmath>
7 using namespace std;
8
9 const int N=250;
10 const int SZ=26;
11 const double eps=1e-8;
12 double p[27],A[N][N];
13 bool inf[N];
14 struct AC
15 {
16 int ch[N][26];
17 int sz,val[N],fail[N];
18 void init(){
19 sz=1;memset(val,0,sizeof(val));
20 memset(ch[0],0,sizeof(ch[0]));
21 }
22 void insert(char *s)
23 {
24 int rt=0;
25 for (int i=0;s[i];i++){
26 int c=s[i]-‘a‘;
27 if (ch[rt][c]==0){
28 ch[rt][c]=sz;
29 memset(ch[sz],0,sizeof(ch[sz]));
30 sz++;
31 }
32 rt=ch[rt][c];
33 }
34 val[rt]=1;
35 }
36 void getfail()
37 {
38 queue<int>q;
39 for (int i=0;i<SZ;i++)
40 {
41 int c=ch[0][i];
42 if (c){ q.push(c);fail[c]=0; }
43 }
44 while (!q.empty())
45 {
46 int u=q.front();q.pop();
47 for (int i=0;i<SZ;i++)
48 {
49 int c=ch[u][i];
50 if (!c){
51 ch[u][i]=ch[fail[u]][i];
52 }else
53 {
54 int v=fail[u];
55 q.push(c);
56 fail[c]=ch[v][i];
57 val[c]|=val[fail[c]];
58 }
59 }
60 }
61 }
62 }ac;
63
64 void build_matrix()
65 {
66 int i,j;
67 memset(A,0,sizeof(A));
68 for(i=0;i<ac.sz;i++)
69 {
70 if(ac.val[i]){A[i][i]=1;A[i][ac.sz]=0;}
71 else
72 {
73 for(j=0;j<SZ;j++){int v=ac.ch[i][j];A[i][v]+=p[j];}
74 A[i][i]+=-1;A[i][ac.sz]=-1;
75 }
76 }
77 }
78
79 void gauss(int n)
80 {
81 int i,j,k,r;
82 for(i=0;i<n;i++)
83 {
84 r=i;
85 for(j=i+1;j<n;j++)
86 if(fabs(A[j][i])>fabs(A[r][i])) r=j;
87 if(fabs(A[r][i])<eps) continue;
88 if(r!=i) for(j=0;j<=n;j++) swap(A[r][j],A[i][j]);
89 for(k=0;k<n;k++) if(k!=i)
90 for(j=n;j>=i;j--) A[k][j]-=A[k][i]/A[i][i]*A[i][j];
91 }
92 memset(inf,0,sizeof(inf));
93 for(i=ac.sz-1;i>=0;i--){
94 if(fabs(A[i][i])<eps&&fabs(A[i][ac.sz])>eps) inf[i]=1;
95 for(j=i+1;j<ac.sz;j++)
96 if(fabs(A[i][j])>eps&&inf[j]) inf[i]=1;
97 }
98 if(inf[0]) printf("Infinity\n");
99 else printf("%.6lf\n",A[0][ac.sz]/A[0][0]+eps);
100 }
101
102 int main()
103 {
104 //freopen("b.txt","w",stdout);
105 int n,i,j;char s[20];
106 while(~scanf("%d",&n))
107 {
108 ac.init();
109 for(i=0;i<SZ;i++) scanf("%lf",p+i);
110 for(i=0;i<n;i++)
111 {
112 scanf("%s",s);ac.insert(s);
113 }
114 ac.getfail();
115 build_matrix();
116 gauss(ac.sz);
117 }
118 return 0;
119 }
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