832B Petya and Exam

题意：给你两个串，第一个串里面的字母都是good 字母，

第二个串是模式串，里面除了字母还有？和*（只有一个）

？可以替换所有good字母， *可以替换所有坏字母和空格（可以是多个坏字母！！！这点卡了我很久，也不举一个样例。。。）

然后q次询问，每次给你一个串，问你能否匹配成功，yes or no

思路：暴力，可惜晚上的时候被hacks掉了，真实数据的范围是超过1e5的，比较可惜。

#include <stdio.h>
#include <string.h>
#include <vector>
#include<math.h>
#include <algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define MAXSIZE  1000005
#define LL long long
using namespace std;

int vis[MAXSIZE];

int solve(char s1[],char s2[],int index)
{
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    if(len1 > len2 + )
        return ;
    int id;
    if(index == -)
    {
        if(len1 != len2)
            return ;
        for(int i=;i<len1;i++)
        {
            if(s1[i] == '?')
            {
                id = s2[i] - 'a';
                if(!vis[id])
                    return ;
            }

            else if(s1[i] != s2[i])
                return ;
        }
        return ;
    }

    else
    {
        int pos1,pos2;
        if(len1 - len2 == )
        {
            pos1 = ;
            pos2 = ;
            while(pos1<len1 && pos2<len2)
            {
                if(s1[pos1] == '*')
                {
                    pos1++;
                    continue;
                }
                if(s1[pos1] == '?')
                {
                    int id = s2[pos2] - 'a';
                    if(!vis[id])
                        return ;
                }
                else if(s1[pos1] != s2[pos2])
                {
                    return ;
                }
                pos1++;
                pos2++;
            }
            return ;
        }
        int s=-,e=-;
        for(int i1=,i2=;i1<len1 && i2<len2;i1++,i2++)
        {
            id = s2[i2] - 'a';
            if(!vis[id] && (s1[i1] != s2[i1]))
            {
                s = i2;
                break;
            }
        }

        for(int i1=len1-,i2=len2-;i1>= && i2>=;i1--,i2--)
        {
            id = s2[i2] - 'a';
            if(!vis[id] && (s1[i1] != s2[i2]))
            {
                e = i2;
                break;
            }
        }

        if(s==- || e==-)
            return ;

        pos1 = ;
        pos2 = ;
        while(pos1<index && pos2<s)
        {
            if(s1[pos1] == '?')
            {
                pos1++;
                pos2++;
                continue;
            }

            if(s1[pos1] != s2[pos2])
                return ;
            pos1++;
            pos2++;
        }
        if(pos1 != pos2 || pos1!=s)
            return ;
        pos1++;
        while(pos2<=e)
        {
            id = s2[pos2] - 'a';
            if(vis[id])
                return ;
            pos2++;
        }
        while(pos1<len1 && pos2<len2)
        {
            if(s1[pos1] == '?')
            {
                pos1++;
                pos2++;
                continue;
            }
            else if(s1[pos1] != s2[pos2])
                return ;
            pos1++;
            pos2++;
        }
        if(pos1!=len1 || pos2!=len2)
            return ;
        return ;
    }
}

int main()
{
    int n,len,index=-;
    char s1[MAXSIZE],s2[MAXSIZE];
    cin >> s1;
    len = strlen(s1);
    for(int i=;i<len;i++)
    {
        int id = s1[i] - 'a';
        vis[id] = ;
    }
    cin >> s1;
    len = strlen(s1);
    for(int i=;i<len;i++)
    {
        if(s1[i] == '*')
        {
            index = i;
            break;
        }
    }
    scanf("%d",&n);
    while(n--)
    {
        cin >> s2;
        int ok = solve(s1,s2,index);
        if(ok)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return ;
}