BZOJ5337 [TJOI2018] 碱基序列 【哈希】【动态规划】

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn = 10200;
 5 
 6 const int mod = 1000000007;
 7 
 8 int n;
 9 string str;
10 string fm[102][12];
11 int num[102];
12 
13 const int Kh = 2;
14 
15 struct hnum{
16     int base;
17     unsigned long long multi[10200];
18     unsigned long long hash[10200];
19 }h[2];
20 
21 void read(){
22     ios::sync_with_stdio(false);
23     cin.tie();
24     cin >> n;
25     cin >> str;
26     for(int i=1;i<=n;i++){
27     cin >> num[i];
28     for(int j=1;j<=num[i];j++) {cin >> fm[i][j];}
29     }
30 }
31 
32 long long rm[2][maxn];
33 int d[maxn];
34 
35 void KMP(int c,int m){
36     memset(d,0,sizeof(d));
37     string &p = fm[c][m];
38     for(int o = 0;o<Kh;o++){
39     unsigned long long now = 0;
40     for(int i=0;i<p.length();i++)
41         now += (p[i]-‘A‘+1)*h[o].multi[i];
42     for(int i=p.length()-1;i<str.length();i++){
43         unsigned long long data;
44         if(i!=p.length()-1)data = h[o].hash[i]-h[o].hash[i-p.length()];
45         else data = h[o].hash[i];
46         if(now == data) d[i]++;
47         now = now*h[o].base;
48     }
49     }
50     for(int i=0;i<str.length();i++) if(d[i]==Kh)d[i]=1;else d[i]=0;
51 }
52 
53 void build_hash(){
54     for(int i=0;i<Kh;i++){
55     h[i].multi[0] = 1;
56     for(int j=1;j<str.length();j++){
57         h[i].multi[j] = h[i].multi[j-1]*h[i].base;
58     }
59     h[i].hash[0] = str[0]-‘A‘+1;
60     for(int j=1;j<str.length();j++){
61         h[i].hash[j] = h[i].hash[j-1]+(str[j]-‘A‘+1)*h[i].multi[j];
62     }
63     }
64 }
65 
66 void work(){
67     h[0].base = 19,h[1].base = 31;
68     build_hash();
69     for(int i=1;i<=num[1];i++){
70     KMP(1,i); for(int k=0;k<str.length();k++) rm[1][k] += d[k];
71     }
72     for(int i=2;i<=n;i++){
73     memset(rm[i&1],0,sizeof(rm[i&1]));
74     for(int j=1;j<=num[i];j++){
75         KMP(i,j); int len = fm[i][j].length();
76         for(int k=0;k<str.length();k++){
77         if(d[k]){
78             if(k-len<0) continue;
79             rm[i&1][k] += rm[(i+1)&1][k-len];
80             rm[i&1][k] %= mod;
81         }
82         }
83     }
84     }
85     long long ans = 0;
86     for(int j=0;j<str.length();j++){
87     ans += rm[n&1][j]; ans %= mod;
88     }
89     printf("%lld",ans);
90 }
91 
92 int main(){
93     read();
94     work();
95     return 0;
96 }