裸的AC自动机，
这倒题不能使用静态数组模拟建树的过程，10000*50*26这样会爆内存，所以使用指针，使用结构体动态分配new
这个可以用来做模板了
1 //#pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 10000 + 100; 36 const int MAXM = 1000010; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 40 const int SIGMA_SIZE = 26; 41 const int MAX_LEN = 52; 42 43 struct Node { 44 int val; 45 Node *ch[SIGMA_SIZE]; 46 Node *f, *last; 47 Node() { 48 rep (i, 0, SIGMA_SIZE - 1) { 49 ch[i] = NULL; 50 } 51 f = last = NULL; 52 val = 0;//非结点 53 } 54 }; 55 56 struct ACMachine { 57 int node_cnt; 58 Node *head; 59 queue<Node*>q; 60 61 void init() { 62 node_cnt = 0; 63 head = new Node(); 64 } 65 int get_idx(char ch) { 66 return ch - 'a'; 67 } 68 void insert_to_tree(char *str) { 69 Node *u = head; 70 for(int i = 0; str[i]; i ++) { 71 int c = get_idx(str[i]); 72 if(!u->ch[c]) { 73 u->ch[c] = new Node(); 74 } 75 u = u->ch[c]; 76 } 77 u->val ++; 78 } 79 void getFail() { 80 q.push(head); 81 while(!q.empty()) { 82 Node *r = q.front(); q.pop(); 83 rep (c, 0, SIGMA_SIZE - 1) { 84 Node *u = r->ch[c]; 85 if(u == NULL) { 86 if(r != head) r->ch[c] = r->f->ch[c]; 87 continue; 88 } 89 q.push(u); 90 if(r == head) { u->f = head; continue; } 91 Node *v = r->f; 92 while(v && v->ch[c] == NULL) v = v->f; 93 u->f = v == NULL ? head : v->ch[c]; 94 u->last = u->f->val ? u->f : u->f->last; 95 } 96 } 97 } 98 int findAns(char *T) { 99 int n = strlen(T), ans = 0; 100 Node *u = head; 101 rep (i, 0, n - 1) { 102 int c = get_idx(T[i]); 103 u = u->ch[c]; 104 if(!u) u = head; 105 Node *v = u; 106 while(v != NULL) { 107 if(v->val >= 0) { 108 ans += v->val; 109 v->val = -1; 110 } 111 else break; 112 v = v->last; 113 } 114 } 115 return ans; 116 } 117 }acm; 118 119 int t, n; 120 char str[MAXM]; 121 122 int main() 123 { 124 // FIN; 125 cin >> t; 126 while(t--) { 127 acm.init(); 128 scanf("%d%*c", &n); 129 rep (i, 1, n) { 130 scanf("%s", str); 131 acm.insert_to_tree(str); 132 } 133 scanf("%s", str); 134 acm.getFail(); 135 cout << acm.findAns(str) << endl; 136 } 137 return 0; 138 }