Bzoj2628 JZPSTR

Solution

Shift-And算法学习笔记

查询复杂度只与模式串相关的字符串匹配算法,考虑 Shift-And 算法。

然后维护每个字符在文本串的出现情况。

Code

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#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define maxn
#define put() putchar('\n')
#define Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
using namespace std;
inline void read(int &x){
int f=1;x=0;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
x*=f;
}
namespace Debug{
Tp void _debug(char* f,Ty t){cerr<<f<<'='<<t<<endl;}
Ts void _debug(char* f,Ty x,Ar... y){while(*f!=',') cerr<<*f++;cerr<<'='<<x<<",";_debug(f+1,y...);}
Tp ostream& operator<<(ostream& os,vector<Ty>& V){os<<"[";for(auto& vv:V) os<<vv<<",";os<<"]";return os;}
#define gdb(...) _debug((char*)#__VA_ARGS__,__VA_ARGS__)
}using namespace Debug;
bitset<1000005>B[10],tmp,cl;
char s[1000005];
signed main(void){
int T,op,l,r,i,len;
read(T);
while (T--) {
read(op);read(l);
if (op==0) {
scanf("%s",s);for (len=0;s[len];len++) ;
for (i=0;i<10;i++) {
tmp=((B[i]>>l)<<l),B[i]=(tmp<<len)^(tmp^B[i]);
}
for (i=0;i<len;i++) B[s[i]-'0'][i+l]=1;
}
else if (op==1){
read(r);
for (i=0;i<10;i++) {
tmp=(B[i]>>r)<<r;
B[i]=(B[i]^((B[i]>>l)<<l))^(tmp>>(r-l));
}
}
else if (op==2) {
read(r);scanf("%s",s);for (len=0;s[len];len++) ;
if (r-l<len) {puts("0");continue;}
tmp=(((~cl)>>l)<<l)^(((~cl)>>(r-len+1))<<(r-len+1));
for (i=0;i<len;i++) {
tmp=tmp&B[s[i]-'0'];
tmp<<=1;
}
printf("%d\n",(int)tmp.count());
}
}
return 0;
}