解码之后是跟普通的自动机求解一下的,只不过解码比较恶心,512=》N》=0 ,所以不能用字符串来存,需要转换成整数来做。
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 53769 12 #define NN 15010 13 #define LL long long 14 #define INF 0xfffffff 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 const int child_num = 258; 19 char s[NN],vir[NN]; 20 int di[NN*8],od[NN],vv[NN]; 21 bool f[N]; 22 class ACAutomation 23 { 24 private: 25 int ch[N][child_num]; 26 int val[N]; 27 int fail[N]; 28 int Q[N]; 29 int id[258]; 30 int sz; 31 // int dp[2][N][1<<10]; 32 public: 33 void init() 34 { 35 fail[0] = 0; 36 for(int i = 0 ; i < child_num; i++) 37 id[i] = i; 38 } 39 void reset()//初始化 40 { 41 memset(ch[0],0,sizeof(ch[0])); 42 memset(val,0,sizeof(val)); 43 sz = 1; 44 } 45 void insert(int *a,int key,int k)//建立trie树 46 { 47 int p = 0; 48 for( int i = 0; i < k ; i++) 49 { 50 int c = id[a[i]]; 51 if(ch[p][c]==0) 52 { 53 memset(ch[sz],0,sizeof(ch[sz])); 54 ch[p][c] = sz++; 55 } 56 p = ch[p][c]; 57 } 58 val[p] += key; 59 } 60 void construct()//构建fail指针 61 { 62 int head = 0,tail = 0,i; 63 for(i = 0 ;i < child_num ; i++) 64 { 65 if(ch[0][i]) 66 { 67 fail[ch[0][i]] = 0; 68 Q[tail++] = ch[0][i]; 69 } 70 } 71 while(head!=tail) 72 { 73 int u = Q[head++]; 74 for(i = 0 ;i < child_num ;i ++) 75 { 76 if(ch[u][i]!=0) 77 { 78 Q[tail++] = ch[u][i]; 79 fail[ch[u][i]] = ch[fail[u]][i]; 80 } 81 else 82 ch[u][i] = ch[fail[u]][i]; 83 } 84 } 85 } 86 int work(int *s,int k) 87 { 88 int p = 0,ans = 0; 89 memset(f,0,sizeof(f)); 90 for(int i = 0; i < k ; i++) 91 { 92 int d = id[s[i]]; 93 p = ch[p][d]; 94 int tmp = p; 95 while(tmp!=0&&!f[tmp]) 96 { 97 ans+=val[tmp]; 98 f[tmp] = 1; 99 tmp = fail[tmp]; 100 } 101 } 102 return ans; 103 } 104 }ac; 105 int change(char *a) 106 { 107 int i,k = strlen(a),j; 108 int g = 0,gg,num =0 ; 109 for(i = 0;i < k ; i++) 110 { 111 if(a[i]=='=') 112 { 113 num++; 114 continue; 115 } 116 int x = od[a[i]];gg=0; 117 for(j=5;j>=0;j--) 118 { 119 di[g++]=((x&(1<<j))>0); 120 } 121 } 122 if(num==1) 123 g-=2; 124 else if(num==2) 125 g-=4; 126 gg = 0; 127 int y = 0; 128 for(j = 0 ;j < g ; j++) 129 { 130 y+=di[j]*(1<<(7-j%8)); 131 if((j+1)%8==0) 132 { 133 vv[gg++] = y; 134 y = 0; 135 } 136 } 137 return g/8; 138 } 139 int main() 140 { 141 int n,i,m; 142 for(i = 'A' ; i <= 'Z' ; i++) 143 od[i] = i-'A'; 144 for(i = 'a' ; i <= 'z' ; i++) 145 od[i] = 26+i-'a'; 146 for(i = '0'; i <= '9' ; i++) 147 od[i] = 52+i-'0'; 148 od['+'] = 62,od['/'] = 63; 149 ac.init(); 150 while(scanf("%d",&n)!=EOF) 151 { 152 ac.reset(); 153 for(i = 1; i <= n; i++) 154 { 155 scanf("%s",vir); 156 int len = change(vir); 157 ac.insert(vv,1,len); 158 } 159 ac.construct(); 160 scanf("%d",&m); 161 for(i = 1; i <= m ;i++) 162 { 163 scanf("%s",s); 164 int len = change(s); 165 printf("%d ",ac.work(vv,len)); 166 } 167 puts(""); 168 } 169 return 0; 170 }