POJ - 1816 Wild Words （模糊狀態下字典樹匹配）

A word is a string of lowercases. A word pattern is a string of lowercases, '?'s and '*'s. In a pattern, a '?' matches any single lowercase, and a '*' matches none or more lowercases.

There are many word patterns and some words in your hand. For each word, your task is to tell which patterns match it.

Input

The first line of input contains two integers N (0 < N <= 100000) and M (0 < M <=100), representing the number of word patterns and the number of words. Each of the following N lines contains a word pattern, assuming all the patterns are numbered from 0 to N-1. After those, each of the last M lines contains a word.

You can assume that the length of patterns will not exceed 6, and the length of words will not exceed 20.

Output

For each word, print a line contains the numbers of matched patterns by increasing order. Each number is followed by a single blank. If there is no pattern that can match the word, print "Not match".

Sample Input

5 4
t*
?h*s
??e*
*s
?*e
this
the
an
is

Sample Output

0 1 3 
0 2 4 
Not match
3

            簡單來說就是字典樹上進行dfs。這裏注意的是，最後保存答案的時候用set是會tle的~~

#include<cstdio>
#include<iostream>
#include<string>
#include<vector>
#include<set>
#include<algorithm>
using namespace std;
const int N = 1e6 + 5;
int nex[N][30];
int tot = 0; vector<int>G[N];
int Hash(char x) {
	if (x == '?')
		return 26;
	if (x == '*')
		return 27;
	return  x - 'a';
}
void insert(string m, int id) {
	int len = m.length();
	int root = 0;
	for (int i = 0; i < len; i++) {
		int id = Hash(m[i]);
		if (nex[root][id] == 0)
			nex[root][id] = ++tot;
		root = nex[root][id];
	}
	G[root].push_back(id);
}
vector<int>ans;
void dfs(string m, int pos, int root) {
	if (pos == m.length()) {
		if (nex[root][27] != 0) {
			for (int i = pos; i <= m.length(); i++) {
				dfs(m, i, nex[root][27]);
			}
		}
		for (int i = 0; i < G[root].size(); i++) {
			ans.push_back(G[root][i]);
		}
		return;
	}
	if (nex[root][Hash(m[pos])] != 0) {
		dfs(m, pos + 1, nex[root][Hash(m[pos])]);
	}
	if (nex[root][26] != 0) {
		dfs(m, pos + 1, nex[root][26]);
	}
	if (nex[root][27] != 0) {
		for (int i = pos; i <= m.length(); i++) {
			dfs(m, i, nex[root][27]);
		}
	}
}
void finstr(string m) {
	ans.clear();
	dfs(m, 0, 0);
	if (ans.size() == 0) {
		cout << "Not match\n";
		return;
	}
	sort(ans.begin(), ans.end()); int sum = 0;
	for (int i = 0; i < ans.size(); i++) {
		if (i == 0 || ans[i] != ans[i - 1]) {
			if (sum != 0)cout << " ";
			cout << ans[i];
			sum++;
		}
	}
	cout << "\n";
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		string t; cin >> t;
		insert(t, i - 1);
	}
	for (int i = 0; i < m; i++) {
		string t; cin >> t;
		finstr(t);
	}
	return 0;
}