POJ 1816 Wild Words
题目大意:
给出N个带通配符(?和*)的模式串, M个询问, 询问一个给你的字符串能匹配哪些模式串. 模式串长度不超过6, 询问串长度不超过20.
简要分析:
带通配符AC自动机? 不是的, 看字符串的长度都那么小, 暴力一下就可以了. 把所有模式串丢到Trie里面, *和?也作为一种转移, 对于每个询问串, 暴力dfs就可以了.
代码实现:
View Code
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #include <vector>
5 #include <algorithm>
6 using namespace std;
7
8 const int MAX_N = 100000, MAX_M = 100, P_LEN = 6, W_LEN = 20;
9 int n, m;
10 char p[P_LEN + 1], w[W_LEN + 1];
11
12 #define pb push_back
13
14 namespace trie {
15 const int MAX_NODE = 200000, SON = 28;
16
17 struct node_t {
18 vector <int> v;
19 node_t *son[SON];
20 node_t() { v.clear(), memset(son, 0, sizeof(son)); }
21 } node_pool[MAX_NODE + 1], *node_idx = node_pool, *root = NULL;
22
23 node_t *node_alloc() {
24 return node_idx ++;
25 }
26
27 void init() {
28 root = node_alloc();
29 }
30
31 void ins(int id, char *str) {
32 node_t *pos = root;
33 while (*str) {
34 int t = *(str ++);
35 t = (t == '?' ? 26 : (t == '*' ? 27 : t - 'a'));
36 if (!pos -> son[t]) pos -> son[t] = node_alloc();
37 pos = pos -> son[t];
38 }
39 pos -> v.pb(id);
40 }
41
42 vector <int> ans;
43 int sz;
44
45 void dfs(char *str, node_t *pos, int idx) {
46 if (str[idx] != 0) {
47 int t = str[idx] - 'a';
48 if (pos -> son[t]) dfs(str, pos -> son[t], idx + 1);
49 if (pos -> son[26]) dfs(str, pos -> son[26], idx + 1);
50 if (pos -> son[27])
51 for (int i = idx; i <= sz; i ++) dfs(str, pos -> son[27], i);
52 }
53 else {
54 for (int i = 0, rb = pos -> v.size(); i < rb; i ++) ans.pb(pos -> v[i]);
55 if (pos -> son[27]) dfs(str, pos -> son[27], idx);
56 }
57 }
58
59 void go(char *str) {
60 ans.clear();
61 sz = strlen(str);
62 dfs(str, root, 0);
63 sort(ans.begin(), ans.end());
64 ans.resize(distance(ans.begin(), unique(ans.begin(), ans.end())));
65 if (!ans.size()) printf("Not match\n");
66 else {
67 for (int i = 0, rb = ans.size(); i < rb; i ++) printf("%d ", ans[i]);
68 printf("\n");
69 }
70 }
71 }
72
73 int main(){
74 scanf("%d%d", &n, &m);
75 trie::init();
76 for (int i = 0; i < n; i ++) {
77 scanf("%s", p);
78 trie::ins(i, p);
79 }
80 for (int i = 0; i < m; i ++) {
81 scanf("%s", w);
82 trie::go(w);
83 }
84 return 0;
85 }
