UVA 11468AC自动机+DP

Posted 2020-08-16 Dirge

dp[i][j]表示走了i步走到j结点的概率。初始值dp[0][0] = 1.当走到的结点不是单词尾结点时，才能走过去。

!end[i]&&last[i] == root时，该结点才可行。

 1 #include <bits/stdc++.h>    
 2 using namespace std;
 3 const int N  = 505;
 4 int id(char c){
 5     if(c >= ‘0‘&&c <= ‘9‘) return c-‘0‘;
 6     if(c >= ‘a‘&&c <= ‘z‘) return c-‘a‘+10;
 7     if(c >= ‘A‘&&c <= ‘Z‘) return c-‘A‘+36;
 8     return -1;
 9 }
10 struct Tire{
11     int nex[N][62], fail[N],end[N], last[N];
12     int root, L;
13     int newnode(){
14         memset(nex[L], -1, sizeof(nex[L]));
15         end[L] = 0;
16         return L++;
17     }
18     void init(){
19         L = 0;
20         root = newnode();
21     }
22     void insert(char* s){
23         int now = root;
24         for(int i = 0; s[i]; i++){
25             int p = id(s[i]);
26             if(nex[now][p] == -1)
27                 nex[now][p] = newnode();
28             now = nex[now][p];
29         }
30         end[now] = 1;
31     }
32     void build(){
33         queue<int> Q;
34         fail[root] = last[root] = root;
35         for(int i = 0; i < 62; i++){
36             int& u = nex[root][i];
37             if(u == -1)
38                 u = root;
39             else{
40                 fail[u] = last[u] = root;
41                 Q.push(u);
42             }
43         }
44         while(!Q.empty()){
45             int now = Q.front();
46             Q.pop();
47             for(int i = 0; i < 62; i++){
48                 int& u = nex[now][i];
49                 if(u == -1)
50                     u = nex[ fail[now] ][i];
51                 else{
52                     fail[u] = nex[ fail[now] ][i];
53                     last[u] = end[ fail[u] ]? fail[u] : last[ fail[u] ];
54                     Q.push(u);
55                 }
56             }
57         }
58     }
59 };
60 Tire ac;
61 double por[70];
62 char s[50];
63 double dp[105][N];
64 int main(){
65     int n, k, t, ca = 1; scanf("%d", &t);
66     while(t--){
67         ac.init();
68         scanf("%d", &k);
69         for(int i = 0; i < k; i++){
70             scanf("%s", s);
71             ac.insert(s);
72         }
73         ac.build();
74         memset(por, 0, sizeof(por));
75         scanf("%d", &n);
76         char ch;
77         for(int i = 0; i < n; i++){
78             scanf(" %c", &ch);
79             int p = id(ch);
80             scanf("%lf", por+p);
81         }
82         int l; scanf("%d", &l);
83         memset(dp, 0, sizeof(dp));
84         dp[0][0] = 1;
85         for(int i = 0; i < l; i++){
86             for(int j = 0; j < ac.L; j++)
87                 if(dp[i][j] > 0) for(int k = 0; k < 62; k++){
88                     int ret = ac.nex[j][k];
89                     if(!ac.end[ret]&&ac.last[ret] == ac.root)
90                         dp[i+1][ret] += dp[i][j]*por[k];
91                 }
92         }
93         double ans = 0;
94         for(int i = 0; i < ac.L; i++)
95             ans += dp[l][i];
96         printf("Case #%d: %f\n", ca++, ans);
97     }
98     return 0;
99 }