第一次做这种题目，参考了下题解，相当于把树扯直了做DP，估计这一类题都是这个套路吧。
状态方程dp[i][next] = dp[i][next]+dp[i][j] ;dp[i][j]表示长度为i的第J个结点的时候满足题意的num，next为当前j点所能走到的下一个合法的结点。
需要用高精度，看到一些规范的高精度写法，觉得不错，有空整理下来。
不知道是不是我理解错了，按理说字符串病毒长度不应超过10.。但开到55依旧RE，开550AC。。。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 110 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 const int child_num = 110; 18 const int BASE = 10000; 19 const int DIG = 4; 20 char s[N*100],vir[550]; 21 int id[2024]; 22 struct bignum 23 { 24 int a[110],len; 25 bignum() 26 { 27 memset(a,0,sizeof(a)); 28 len = 1; 29 } 30 bignum(int v) 31 { 32 memset(a,0,sizeof(a)); 33 len = 0; 34 do 35 { 36 a[len++] = v%BASE; 37 v/=BASE; 38 }while(v); 39 } 40 /*bignum(const char s[]) 41 { 42 memset(a,0,sizeof(a)); 43 int k = strlen(s); 44 len = k/DIG; 45 if(k%DIG) len++; 46 int cnt = 0; 47 for(int i = k-1; i >= 0 ; i-=DIG) 48 { 49 int t = 0; 50 int kk = i-DIG+1; 51 if(kk<0) kk =0; 52 for(int j = kk ; j <= i ; j++) 53 t = t*10+s[j]-'0'; 54 a[cnt++] = t; 55 } 56 }*/ 57 bignum operator + (const bignum &b)const 58 { 59 bignum res; 60 res.len = max(len,b.len); 61 int i; 62 for(i = 0 ; i < res.len ;i ++) 63 res.a[i] = 0; 64 for(i = 0 ; i < res.len ; i++) 65 { 66 res.a[i] += ((i<len)?a[i]:0)+((i<b.len)?b.a[i]:0); 67 res.a[i+1] += res.a[i]/BASE; 68 res.a[i] = res.a[i]%BASE; 69 } 70 if(res.a[res.len]>0) res.len++; 71 return res; 72 } 73 void output() 74 { 75 printf("%d",a[len-1]); 76 for(int i = len-2 ; i >=0 ; i--) 77 printf("%04d",a[i]); 78 printf(" "); 79 } 80 }dp[110][110]; 81 class AC 82 { 83 private: 84 int ch[N][child_num]; 85 int Q[N]; 86 int val[N]; 87 int fail[N]; 88 //int id[N]; 89 int sz; 90 public : 91 void init() 92 { 93 fail[0] = 0; 94 //for(int i = 0 ;i < child_num-32 ; i++) 95 //id[i+32] = i; 96 } 97 void reset() 98 { 99 memset(val,0,sizeof(val)); 100 memset(fail,0,sizeof(fail)); 101 memset(ch[0],0,sizeof(ch[0])); 102 sz = 1; 103 } 104 void insert(char *a,int key) 105 { 106 int k = strlen(a),p = 0; 107 for(int i = 0 ; i < k ;i++) 108 { 109 int d = id[a[i]]; 110 if(ch[p][d]==0) 111 { 112 memset(ch[sz],0,sizeof(ch[sz])); 113 ch[p][d] = sz++; 114 } 115 p = ch[p][d]; 116 } 117 val[p] = key; 118 } 119 void construct(int n) 120 { 121 int i,head=0,tail = 0; 122 for(i = 0; i < n ; i++) 123 { 124 if(ch[0][i]) 125 { 126 Q[tail++] = ch[0][i]; 127 fail[ch[0][i]] = 0; 128 } 129 } 130 while(head!=tail) 131 { 132 int u = Q[head++]; 133 val[u]|=val[fail[u]]; 134 for(i = 0 ; i < n ; i++) 135 { 136 if(ch[u][i]) 137 { 138 Q[tail++] = ch[u][i]; 139 fail[ch[u][i]] = ch[fail[u]][i]; 140 } 141 else ch[u][i] = ch[fail[u]][i]; 142 } 143 } 144 } 145 void work(int m,int n) 146 { 147 int i,j,g; 148 for(i = 1; i <= m ;i++) 149 for(j = 0 ;j <= sz; j++) 150 dp[i][j] = bignum(0); 151 dp[0][0] = bignum(1); 152 for(i = 0 ; i < m ;i++) 153 { 154 for(j = 0 ; j < sz ;j++) 155 for(g = 0 ; g < n ; g++) 156 if(!val[ch[j][g]]) 157 { 158 dp[i+1][ch[j][g]]=dp[i+1][ch[j][g]]+dp[i][j]; 159 } 160 } 161 bignum ans = bignum(0); 162 for(j = 0 ;j < sz ; j++) 163 ans=ans+dp[m][j]; 164 ans.output(); 165 } 166 }ac; 167 int main() 168 { 169 int n,m,i,p; 170 ac.init(); 171 while(cin>>n>>m>>p) 172 { 173 cin>>s; 174 for(i = 0 ; i < n; i++) 175 id[s[i]] = i; 176 ac.reset(); 177 for(i = 1;i <= p; i++) 178 { 179 scanf("%s",vir); 180 ac.insert(vir,1); 181 } 182 ac.construct(n); 183 ac.work(m,n); 184 } 185 return 0; 186 }