hdu2594 Simpsons’ Hidden Talents LCS--扩展KMP

Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in politics, OK?
Marge: OK.
Homer: So I take some politician’s name, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.

题意：求两个字符串的最长公共子串

扩展KMP裸题

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 using namespace std;

 const int maxn=5e4+;

 char s[maxn],t[maxn];
 int nxt[maxn],ext[maxn];

 void EKMP(char s[],char t[],int lens,int lent){
     int i,j,p,l,k;
     nxt[]=lent;j=;
     while(j+<lent&&t[j]==t[j+])j++;
     nxt[]=j;
     k=;
     for(i=;i<lent;i++){
         p=nxt[k]+k-;
         l=nxt[i-k];
         if(i+l<p+)nxt[i]=l;
         else{
             j=max(,p-i+);
             while(i+j<lent&&t[i+j]==t[j])j++;
             nxt[i]=j;
             k=i;
         }
     }

     j=;
     while(j<lens&&j<lent&&s[j]==t[j])j++;
     ext[]=j;k=;
     for(i=;i<lens;i++){
         p=ext[k]+k-;
         l=nxt[i-k];
         if(l+i<p+)ext[i]=l;
         else{
             j=max(,p-i+);
             while(i+j<lens&&j<lent&&s[i+j]==t[j])j++;
             ext[i]=j;
             k=i;
         }
     }
 }

 int main(){
     while(scanf("%s%s",s,t)!=EOF){
         EKMP(t,s,strlen(t),strlen(s));
         int l=strlen(t);
         int i;
         for(i=;i<l;++i){
             if(ext[i]==l-i){
                 printf("%s %d\n",t+i,l-i);
                 break;
             }
         }
         if(i==l)printf("0\n");
     }
     return ;
 }