POJ 1816 - Wild Words - [字典树+DFS]

Posted 2021-01-20 dilthey

题目链接：

http://poj.org/problem?id=1816

http://bailian.openjudge.cn/practice/1816?lang=en_US

Time Limit: 2000MS　　Memory Limit: 65536K

Description

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

 

题意：

给出 $n$ 个模式串（只包含小写字母，"?" 和 "*"），$m$ 个字符串（只包含小写字母），

模式串中 "?" 代表能匹配任意一个字母，"*" 代表能匹配任意多个字母（可以是 $0$ 个）。

现在对于 $m$ 个字符串，查询编号为 $0 sim n-1$ 这 $n$ 个模式串中有多少个是可以匹配的上的。

 

题解：

对 $n$ 个模式串建立字典树；对于 $m$ 个字符串的查询，每个字符串都在字典树上进行DFS匹配。

放几组数据：

3 1
t
t*
t**
t

1 1
*j*j
jjj

2 1
abc
abc
abc

 

AC代码：

#include<cstdio>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=100000+5;
int n,m;

namespace Trie
{
    const int SIZE=maxn;
    int sz;
    struct TrieNode{
        vector<int> pos;
        int nxt[28];
    }trie[SIZE];
    void init(){sz=1;}
    int idx(const char& c)
    {
        if(c==‘?‘) return 26;
        if(c==‘*‘) return 27;
        return c-‘a‘;
    }
    void insert(const string& s,int id)
    {
        int p=1;
        for(int i=0;i<s.size();i++)
        {
            int ch=idx(s[i]);
            if(!trie[p].nxt[ch]) trie[p].nxt[ch]=++sz;
            p=trie[p].nxt[ch];
        }
        trie[p].pos.push_back(id);
    }

    void dfs(vector<int>& ans,int p,const string& s,int k)
    {
        if(trie[p].nxt[27]) {
            for(int i=0;k+i<=s.size();i++) dfs(ans,trie[p].nxt[27],s,k+i);
        }
        if(k>=s.size())
        {
            for(int i=0;i<trie[p].pos.size();i++) ans.push_back(trie[p].pos[i]);
            return;
        }
        int ch=idx(s[k]);
        if(trie[p].nxt[26]) dfs(ans,trie[p].nxt[26],s,k+1);
        if(trie[p].nxt[ch]) dfs(ans,trie[p].nxt[ch],s,k+1);
    }
};

int main()
{
    scanf("%d%d",&n,&m);
    Trie::init();
    char s[23];
    for(int i=0;i<n;i++)
    {
        scanf("%s",&s);
        Trie::insert(s,i);
    }

    vector<int> ans;
    for(int i=1;i<=m;i++)
    {
        scanf("%s",&s);
        ans.clear();
        Trie::dfs(ans,1,s,0);
        if(ans.empty()) printf("Not match
");
        else
        {
            sort(ans.begin(),ans.end());
            ans.erase(unique(ans.begin(),ans.end()),ans.end());
            for(int k=0;k<ans.size();k++) {
                printf("%d%c",ans[k],(k==ans.size()-1)?‘
‘:‘ ‘);
            }
        }
    }
}