[bzoj3507 Cqoi2014]通配符匹配 (hash+DP)

传送门

Solution

显然用哈希233
设(f[i][j])表示第i个通配符和当前第j个字符是否匹配
考虑两种通配符的特性，直接转移即可

Code

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
using namespace std;
typedef unsigned long long ull;
 
const int N=200005,bas=233;
int n,len,tot;
bool f[20][N];
int cl[20];
ull da[N],hs[N],pw[N];
char s[N],ch[N];
 
int main() {
    pw[0]=1;F(i,1,N-1) pw[i]=pw[i-1]*bas;
    scanf("%s",s+1);len=strlen(s+1);s[++len]='?';
    F(i,1,len) {
        if(s[i]=='*'||s[i]=='?') cl[++tot]=i;
        da[i]=da[i-1]*bas+s[i];
    }
    scanf("%d",&n);
    while(n--) {
        memset(f,0,sizeof(f));f[0][0]=1;
        scanf("%s",ch+1);len=strlen(ch+1);ch[++len]='~';
        F(i,1,len) hs[i]=hs[i-1]*bas+ch[i];
        F(i,0,tot) {
            if(s[cl[i]]=='*') F(j,1,len) f[i][j]|=f[i][j-1];
            F(j,0,len) if(f[i][j]) {
                int l1=cl[i]+1,r1=cl[i+1]-1;
                int l2=j+1,r2=j+cl[i+1]-cl[i]-1;
//              cout<<l1<<" "<<r1<<" "<<l2<<" "<<r2<<endl;
                if(da[r1]-da[l1-1]*pw[r1-l1+1]!=hs[r2]-hs[l2-1]*pw[r2-l2+1]) continue;
                if(s[cl[i+1]]=='?') f[i+1][r2+1]=1;//,cout<<i+1<<" "<<r2+1<<endl;
                else f[i+1][r2]=1;//,cout<<i+1<<" "<<r2<<endl;
            }
        }
        if(f[tot][len]) puts("YES");
        else puts("NO");
    }
    return 0;
}