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  • 【CQOI2014】通配符匹配

    题面

    https://www.luogu.org/problem/P3167

    题解

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<vector>
    #define N 100500
    #define ri register int
    #define uLL unsigned long long
    
    const uLL p=107;
    using namespace std;
    uLL pp[N];
    struct hash {
      uLL sum[N];
      inline void getsum(char *s) {
        sum[0]=0;
        int l=strlen(s+1);
        for (ri i=1;i<=l;i++) sum[i]=sum[i-1]*p+(s[i]-'a');
      }
      inline uLL getval(int l,int r) {
        //if (l>r) return 0;
        return sum[r]-sum[l-1]*pp[r-l+1];
      }
    } H1,H2;
    
    bool f[15][N];
    char s1[N],s2[N];
    int L[15],loc[15];
    
    int main(){
      pp[0]=1;
      for (ri i=1;i<N;i++) pp[i]=pp[i-1]*p;
      scanf("%s",s1+1);
      int l=strlen(s1+1);
      s1[l+2]=s1[l+1]; s1[l+1]='?';
      int cnt=0,pre=0;
      loc[0]=0;
      for (ri i=1;i<=l+1;i++) if (s1[i]=='?'||s1[i]=='*') {
        loc[++cnt]=i;
        L[cnt]=i-pre-1;
        pre=i;
      }
      H1.getsum(s1);
      int n;
      scanf("%d",&n);
      while (n--) {
        scanf("%s",s2+1);
        int l=strlen(s2+1);
        s2[l+2]=s2[l+1]; s2[l+1]='#';
        H2.getsum(s2);
        memset(f,0,sizeof(f));
        f[0][0]=1;
        for (ri i=1;i<=cnt;i++) 
          for (ri j=0;j<=l+1;j++) {
            if (s1[loc[i]]=='*') {
              if (H1.getval(loc[i-1]+1,loc[i]-1)==H2.getval(j-L[i]+1,j)&&f[i-1][j-L[i]]) f[i][j]=1;
              if (H1.getval(loc[i-1]+1,loc[i]-1)==H2.getval(j-L[i],j-1)&&f[i-1][j-L[i]-1]) f[i][j]=1;
              f[i][j]|=f[i][j-1];
            }
            else {
              if (H1.getval(loc[i-1]+1,loc[i]-1)==H2.getval(j-L[i],j-1)&&f[i-1][j-L[i]-1]) f[i][j]=1;
            }
          }
        if (f[cnt][l+1]==1) puts("YES"); else puts("NO");
      }
    }
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  • 原文地址:https://www.cnblogs.com/shxnb666/p/11279812.html
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