luogu problem solving P3763 [[TJOI2017]DNA]

• Topic link:

https://www.luogu.org/problemnew/show/P3763

• Ideas:

First we're going to use the Rabin-Karp hash, which is actually this:

若\(w_{str}\)=(\(a_0\) \(p^{n-1}\)+\(a_1\) \(p^{n-2}\)+...+\(a_{n-1}\) \(p^0\))

so

\(w_{pre_{i-1}}\) \(=(\) \(a_0\) \(p^{i-1}\)+\(a_1\) \(p^{i-2}\)+...+\(a_{i-1}\) \(p^0\))

\(w_{pre_{j}}\) \(=(\) \(a_0\) \(p^{j}\)+\(a_1\) \(p^{j-1}\)+...+\(a_{j}\) \(p^0\))

so

\(w_{str_{i,j}}\)

\(=(\) \(a_i\) \(p^{j-i}\)+\(a_{i+1}\) \(p^{j-i-1}\)+...+\(a_{j}\) \(p^0\))

\(=\) \(w_{pre_{j}}\) \(-\) \(w_{pre_{i-1}}\) \(p^{j-i+1}\)

  注意了，我这里并没有取模，而是直接直接用unsigned long long 自然溢出，这样更快

However, most people use dichotomy to determine the right endpoint, and I figured out a tricky operation myself (as if I've heard it somewhere) ---- use multiplication .

Because in this question, I think the upper and lower bounds of the two points may be quite different. Although theoretically the average of the two points is better, but in this question, it is actually measured to use the doubling faster (if someone uses the two points to run later than I'm ignoring my words soon).

The idea is this, see the code for other details.

• Code:

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iostream>
#define ri int 
const int maxn=100005;
typedef long long   ll;
typedef unsigned long long ull;
char a[maxn],b[maxn];
ull q[maxn],aht[maxn],bht[maxn];
int cnt=0,lena,lenb;
inline void Hash(){
    ll x=0;
    for(ri i=0;a[i];i++){
        x=x*131+a[i]-31;
        aht[i]=x;
    //  cout<<x<<i<<endl;
    }x=0;
    for(ri i=0;b[i];i++){
        x=x*131+b[i]-31;
        bht[i]=x;
    }x=0;
    return ;
}
inline int solve(int x,int y){
    int k=0,p=1;
    x++,y++;
    while(p!=0){  
        if((aht[x+k+p-1]-aht[x-1-1]*q[k+p+1])==(bht[y+k+p-1]-bht[y-1-1]*q[k+p+1]))k+=p,p*=2;
        else p=p/2;//cout<<l<<'*'<<r<<'*'<<mid<<endl;
        while(x+k+p>lena||y+k+p>lenb)p=p/2;
    }
    if(a[x-1]==b[y-1])k++;
    return k;
}
inline bool ok(int i){
      int la=0,k;
      for(ri j=1;j<=3;j++){
           k=solve(i,la);
           i+=k+1,la+=k+1;
           if(la>=lenb){return 1;}//cout<<i<<' '<<la<<endl;
      }
         k=solve(i,la); 
         i+=k;la+=k;
      //   cout<<i<<' '<<la<<endl;
         if(la>=lenb) return 1;
         return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    q[0]=1;
    for(ri i=1;i<=100001;i++)
        q[i]=(ull)q[i-1]*131;
  //预处理幂，这个技巧来自https://www.cnblogs.com/sineagle/p/8490655.html
    while(t--){
       int cnt=0;
       scanf("%s",a);
       scanf("%s",b);
       lena=strlen(a),lenb=strlen(b);
       if(lena<lenb){printf("0\n");continue;}
       Hash();
       for(ri i=0;i<lena-lenb+1;i++){
         if(ok(i))cnt++;
       }
       printf("%d\n",cnt);
       memset(aht,0,sizeof(aht));
       memset(bht,0,sizeof(bht));
    }
    return 0;
}

• postscript:

The fastest run on luogu was 240ms, but it was still no match for BZOJ's dalao