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hdu 3992 AC自动机上的高斯消元求期望

2024-07-17 20:18:40   218人阅读

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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