POJ

Wild Words POJ - 1816

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

不用看都知道是字典树（刷的字典树专题，嘿嘿），建立字典树，然后DFS爆搜就行，刚开始一直TLE，后来发现是SET的事

题意：上边给N个类似正则表达式的玩意，‘*’表示匹配0个或多个字母，‘？’表示一个，问有哪几种是能匹配到的，全部不能NOT MATCH

#include<cstdio>#include<iostream>#include<cmath>#include<queue>#include<vector>#include<cstring>#include<algorithm>#include<map>#include<set>#include<stack>#include<sstream>#include<string>using namespace std;#define inf 0x3f3f3f3fstruct node{    node* next[28];    int mos[101];    int ges; //  vector<int>q; }ss[100000];int o;inline node *add(){    node *t=&ss[o++];    memset(t->next,0,sizeof(t->next));    t->ges=0;//  t->q.clear();    return t;}void inse(string a,node *root,int pp){    int len=a.length();    for(int i=0;i<len;i++)    {        if(a[i]>='a'&&a[i]<='z')        {               int tt=a[i]-'a';            if(root->next[tt])            {                    root=root->next[tt];            }            else            {                node *t=add();            //  node *t=&ss[o++];            //  memset(t->next,0,sizeof(t->next));//  t->ges=0;            //  t->q.clear();                root->next[tt]=t;                root=root->next[tt];            }        }        else        {            if(a[i]=='?')            {                if(root->next[26])                {                    root=root->next[26];                }                else                {                node *t=add();            //  node *t=&ss[o++];    //memset(t->next,0,sizeof(t->next));//  t->ges=0;//  t->q.clear();                root->next[26]=t;                root=root->next[26];                }            }            else            {                if(root->next[27])                {                    root=root->next[27];                }                else                {                    node *t=add();            //  node *t=&ss[o++];//  memset(t->next,0,sizeof(t->next));//  t->ges=0;//  t->q.clear();                    root->next[27]=t;                    root=root->next[27];                }            }        }    }    root->mos[(root->ges)++]=pp;//root->q.push_back(pp);    }//set<int>qq;int ans[100000];int ans_sum;void dfs(char a[],node *root){           if(a[0]==0)    {        while(root)        {            int len=root->ges;        //  int len=root->q.size();            for(int i=0;i<len;i++)            {            //qq.insert(root->q[i]);            //ans[ans_sum++]=root->q[i];            ans[ans_sum++]=root->mos[i];            }            root=root->next[27];        }        return;    }    if(root->next[26]!=NULL)    {        dfs(a+1,root->next[26]);    }    if(root->next[27]!=NULL)    {           int len=strlen(a);        for(int i=0;i<=len;i++ )        {            dfs(a+i,root->next[27]);        }    }    if(root->next[a[0]-'a'])    {        dfs(a+1,root->next[a[0]-'a']);    }}int main(){       int n,m;    o=0;    cin>>n>>m;    node *root=add();//  node *root=&ss[o++];//  memset(root->next,0,sizeof(root->next));//  t->ges=0;//  root->q.clear();    //=add();    int pp=0;    while(n--)    {   string a;        cin>>a;        inse(a,root,pp++);    }    char b[1000];    while(m--)    {        //qq.clear();        cin>>b;        ans_sum=0;        dfs(b,root);        if(ans_sum==0)        {            cout<<"Not match"<<endl;            continue;        }        //set<int>::iterator it;        //for(it=qq.begin();it!=qq.end();it++)        //cout<<*it<<" ";        //cout<<endl;        sort(ans,ans+ans_sum);        cout<<ans[0];        for(int i=1;i<ans_sum;i++)        if(ans[i]!=ans[i-1])        cout<<" "<<ans[i];        cout<<endl;    }    return 0;}