bzoj 4084 双旋转字符串

给两个集合A,B,找满足要求的(a,b)的对数,可以计算对于a,哪些b成立.

还有就是字符串hash的使用,感觉平时用字符串hash太少了.

 /**************************************************************
     Problem: 4084
     User: idy002
     Language: C++
     Result: Accepted
     Time:6456 ms
     Memory:290272 kb
 ****************************************************************/
 
 #include <cstdio>
 #include <set>
 #include <vector>
 #include <algorithm>
 #define N 8000010
 #define Base 31
 #define Mod 1000000007
 using namespace std;
 
 typedef long long dnt;
 
 int n, m, ln, lm;
 char *sa[N], *sb[N];
 char buf_arr[N], *buf=buf_arr;
 dnt fx[N], sx[N], pow[N];
 int fail[N];
 multiset<int> stb;
 
 int hash( char *s ) {
     dnt rt = ;
     for( int i=; s[i]; i++ )
         rt = (rt*Base + s[i]-'a') % Mod;
     return rt;
 }
 void init_hash( int ln, char *s ) {     //  ln>=1
     fx[] = s[]-'a';
     for( int j=; j<ln; j++ )
         fx[j] = (fx[j-]*Base + s[j]-'a') % Mod;
     sx[ln-] = s[ln-]-'a';
     for( int j=ln-; j>=; j-- )
         sx[j] = ((s[j]-'a')*pow[ln--j] + sx[j+]) % Mod;
 }
 void init_fail( char *s ) {
     fail[] = ;
     fail[] = ;
     for( int i=; s[i]; i++ ) {
         int j=fail[i];
         while( j && s[j]!=s[i] ) j=fail[j];
         if( s[j]==s[i] )
             fail[i+]=j+;
         else
             fail[i+]=;
     }
 }
 void work1() {  //  ln > lm
     vector<int> vc;
     int ll = (ln+lm)>>;
     dnt ans = ;
     for( int t=; t<=n; t++ ) {
         init_fail(sa[t]+ll);
         for( int i=; i<ll; i++ )
             buf[i] = sa[t][i];
         for( int i=; i<ll; i++ )
             buf[ll+i] = sa[t][i];
         init_hash(ll+ll,buf);
         vc.clear();
         int i=, j=;
         while( i<ln- ) {
             while( i<ln- && j<ln-ll && buf[i]==sa[t][ll+j] ) i++, j++;
             if( j==ln-ll ) {
                 dnt v;
                 v = fx[i+(ll+ll-ln)-]-fx[i-]*pow[ll+ll-ln];
                 v = (v%Mod+Mod) % Mod;
                 vc.push_back(v);
                 j = fail[ln-ll];
             }
             if( i==ln- ) break;
             while( j && sa[t][ll+j]!=buf[i] ) j=fail[j];
             if( sa[t][ll+j]!=buf[i] ) i++;
         }
         sort( vc.begin(), vc.end() );
         vc.erase( unique(vc.begin(),vc.end()), vc.end() );
         for( int t=; t<vc.size(); t++ )
             ans += stb.count(vc[t]);
     }
     printf( "%lld\n", ans );
 }
 void work3() {  //  ln = lm
     vector<int> vc;
     dnt ans = ;
     for( int i=; i<=n; i++ ) {
         init_hash(ln,sa[i]);
         vc.clear();
         vc.push_back( sx[] );
         for( int j=; j<ln; j++ ) {
             int v = ((dnt)sx[j]*pow[j]+fx[j-]) % Mod;
             vc.push_back(v);
         }
         sort( vc.begin(), vc.end() );
         vc.erase( unique(vc.begin(), vc.end()), vc.end() );
         for( int t=; t<vc.size(); t++ )
             ans += stb.count(vc[t]);
     }
     printf( "%lld\n", ans );
 }
 int main() {
     scanf( "%d%d%d%d", &n, &m, &ln, &lm );
     if( ln>lm ) {
         for( int i=; i<=n; i++ ) {
             scanf( "%s", buf );
             sa[i] = buf;
             buf += ln+;
         }
         for( int i=; i<=m; i++ ) {
             scanf( "%s", buf );
             sb[i] = buf;
             buf += lm+;
         }
     } else {
         swap(n,m);
         swap(ln,lm);
         for( int i=; i<=m; i++ ) {
             scanf( "%s", buf );
             sb[i] = buf;
             reverse( buf, buf+lm );
             buf += lm+;
         }
         for( int i=; i<=n; i++ ) {
             scanf( "%s", buf );
             sa[i] = buf;
             reverse( buf, buf+ln );
             buf += ln+;
         }
     }
     pow[] = ;
     for( int i=; i<=ln; i++ )
         pow[i] = pow[i-]*Base % Mod;
     for( int i=; i<=m; i++ )
         stb.insert( hash(sb[i]) );
     if( ln!=lm )
         work1();
     else
         work3();
 }