题意：给你base64编码后的模式串和文本串，让你看编码之前的文本串和分别包含了多少模式串
解题思路：主要是编码还有注意分支要开256 ，然后就是裸的AC自动机
解题代码：
1 // File Name: temp.cpp 2 // Author: darkdream 3 // Created Time: 2014年09月11日 星期四 15时18分256秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 #include<queue> 25 #define LL long long 26 #define maxn 512*65 27 using namespace std; 28 int hs[515]; 29 struct Trie 30 { 31 int next[maxn][256],fail[maxn],end[maxn]; 32 int root, L; 33 int newnode() 34 { 35 memset(next[L],-1,sizeof(next[L])); 36 end[L++] = 0 ; 37 return L-1; 38 } 39 void init() 40 { 41 L = 0 ; 42 root = newnode(); 43 } 44 void insert(int buf[],int len,int K) 45 { 46 int now = root; 47 for(int i = 0 ;i < len ;i ++) 48 { 49 if(next[now][buf[i]] == -1) 50 { 51 next[now][buf[i]] = newnode(); 52 } 53 now = next[now][buf[i]]; 54 } 55 end[now] = K; 56 } 57 void build() 58 { 59 queue<int> Q; 60 fail[root] = root; 61 for(int i = 0;i < 256;i ++) 62 { 63 if(next[root][i] == -1) 64 { 65 next[root][i] = root; //指向root 是没有问题的，我们主要是通过end 数组对个数进行计数的。 66 }else{ 67 fail[next[root][i]] = root; 68 Q.push(next[root][i]); 69 } 70 } 71 while(!Q.empty()) 72 { 73 int now = Q.front(); 74 Q.pop(); 75 for(int i = 0 ;i < 256;i ++) 76 { 77 if(next[now][i] == -1) 78 next[now][i] = next[fail[now]][i]; 79 else{ 80 fail[next[now][i]] = next[fail[now]][i]; 81 Q.push(next[now][i]); 82 } 83 } 84 } 85 } 86 void query(int buf[],int len) 87 { 88 int now = root ; 89 for(int i = 0 ;i < len;i ++) 90 { 91 now = next[now][buf[i]]; 92 int temp = now ; 93 while(temp != root) 94 { 95 if(end[temp]) 96 { 97 hs[end[temp]] = 1; 98 } 99 temp = fail[temp]; 100 } 101 } 102 } 103 }; 104 char buf[3058]; 105 int temp[3058]; 106 Trie ac; 107 int cbit(char c) 108 { 109 if(c >= 'A' && c <= 'Z') 110 return c -'A'; 111 if(c >= 'a' && c <= 'z') 112 return c -'a' + 26; 113 if(c >= '0' && c <= 'z') 114 return c -'0' + 52; 115 if(c == '+') 116 return 62; 117 else if(c == '/')return 63; 118 } 119 void debug(int x) 120 { 121 printf("%d ",x); 122 for(int i = 0 ;i < x; i ++) 123 printf("%c",temp[i]); 124 printf(" "); 125 } 126 int lt = 0; 127 void change(){ 128 129 int len = strlen(buf); 130 int t = 0; 131 int ln = 0 ; 132 int tt; 133 lt = 0; 134 for(int i = 0 ;i < len;i ++) 135 { 136 if(buf[i] != '=') 137 { 138 if(buf[i] >= 'A' && buf[i] <= 'Z') 139 tt = buf[i] -'A'; 140 else if(buf[i] >= 'a' && buf[i] <= 'z') 141 tt = buf[i]-'a' + 26; 142 else if(buf[i] >= '0' && buf[i] <= 'z') 143 tt= buf[i] -'0' + 52; 144 else if(buf[i] == '+') 145 tt = 62; 146 else if(buf[i] == '/')tt = 63; 147 if(6 < 8 - t) 148 { 149 ln += (tt << (8-t-6)); 150 t += 6; 151 }else{ 152 ln += (tt >> (6-8+t)); 153 temp[lt++] = ln; 154 ln = (tt&((1<<(6-8+t)) -1)) << (8-(6-8+t)); 155 t = (6-8+t); 156 } 157 } 158 } 159 //printf("%d %d ",lt,rlen); 160 } 161 162 int main(){ 163 int n; 164 while(scanf("%d",&n) != EOF) 165 { 166 ac.init(); 167 for(int i =1 ;i <= n;i ++) 168 { 169 scanf("%s",buf); 170 change(); 171 ac.insert(temp,lt,i); 172 // debug(lt); 173 } 174 ac.build(); 175 int m; 176 scanf("%d",&m); 177 for(int i = 1;i <= m;i ++) 178 { 179 memset(hs,0,sizeof(hs)); 180 scanf("%s",buf); 181 change(); 182 // debug(lt); 183 int ans = 0; 184 ac.query(temp,lt); 185 for(int i = 1;i <= n;i ++) 186 if(hs[i]) 187 { 188 ans ++ ; 189 } 190 printf("%d ",ans); 191 } 192 printf(" "); 193 } 194 return 0; 195 }