[BZOJ3507][CQOI2014]通配符匹配(DP+Hash)

显然f[i][j]表示S匹配到第i个通配符，T匹配到第j个字符，是否可行。

一次一起转移两个通配符之间的所有字符，Hash判断。

稍微有点细节。常数极大卡时过排名倒数，可能是没自然溢出的原因。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=,p1=,p2=;
 bool f[][N];
 char s1[N],s2[N];
 int n,m,T,tot,pos[],h1[N],h2[N],pw[N];

 int main(){
     freopen("bzoj3507.in","r",stdin);
     freopen("bzoj3507.out","w",stdout);
     scanf("%s",s1+); n=strlen(s1+)+; s1[n]='?';
     pw[]=; rep(i,,n) pw[i]=1ll*pw[i-]*p1%p2;
     rep(i,,n) h1[i]=(1ll*h1[i-]*p1+s1[i])%p2;
     rep(i,,n) if (s1[i]=='?' || s1[i]=='*') pos[++tot]=i;
     for (scanf("%d",&T); T--; ){
         scanf("%s",s2+); m=strlen(s2+)+; s2[m]='#';
         rep(i,,tot) rep(j,,m) f[i][j]=; f[][]=;
         rep(i,,m) h2[i]=(1ll*h2[i-]*p1+s2[i])%p2;
         rep(i,,tot) rep(j,,m){
             if (s1[pos[i]]=='*') f[i][j]|=f[i][j-];
             if (!f[i][j]) continue;
             int l1=pos[i]+,r1=pos[i+]-,l2=j+,r2=j+(pos[i+]-pos[i])-;
             if ((h1[r1]-1ll*h1[l1-]*pw[r1-l1+]%p2+p2)%p2==(h2[r2]-1ll*h2[l2-]*pw[r2-l2+]%p2+p2)%p2){
                 if (s1[pos[i+]]=='?') f[i+][r2+]|=f[i][j]; else f[i+][r2]|=f[i][j];
             }
         }
         puts(f[tot][m] ? "YES" : "NO");
     }
     return ;
 }