最长公共前缀（LCP）

http://vjudge.net/contest/view.action?cid=47286#problem/C

Description

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.

 

Input

Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.

 

Output

Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0. 
The lengths of s1 and s2 will be at most 50000.

 

Sample Input

     
     
     
     

      
      
      
       clinton
homer
riemann
marjorie 
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
       0
rie 3 
     
     
     
     

最长公共前缀。将两字符串拼到一起注意拼的时候在中间加一个不会出现的字符

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef unsigned long long int ULL;
const int SIZE = 100003;
const int SEED = 13331;
const int maxn = 600100;
char a[maxn],b[maxn],s[maxn];
ULL H[maxn];
ULL XL[maxn];
int len;
void build(char *s){
    len=strlen(s);
    H[len]=0;
    XL[0]=1;
    for (int i=len-1;i>=0;i--){
        H[i]=H[i+1]*SEED+s[i];
        XL[len-i]=XL[len-i-1]*SEED;
    }
}
ULL hash(int i,int L){
    return H[i]-H[i+L]*XL[L];
}
int lcp(int i,int j){
    int l=0,r=min(len-i,len-j);
    int res=0;
    while (l<=r){
        int mid=(l+r)/2;
        if (hash(i,mid)==hash(j,mid)){
            res=mid;
            l=mid+1;
        }
        else{
            r=mid-1;
        }
    }
    return res;
}
int main()
{
    while(~scanf("%s%s",a,b))
    {
        memset(s,0,sizeof(s));
        int n=strlen(a);
        int m=strlen(b);
        for(int i=0;i<n;i++)
            s[i]=a[i];
        s[n]='$';
        for(int j=0;j<m;j++)
            s[j+1+n]=b[j];
        len=strlen(s);
        build(s);
        //printf("%s\n",s);
        int flag=1;
        for(int i=n+1;i<len;i++)
        {
            //printf("%d ",lcp(0,i));
            if(lcp(0,i)==len-i)
            {
                for(int j=i;j<len;j++)
                    printf("%c",s[j]);
                printf(" %d\n",len-i);
                flag=0;
                break;
            }
        }
        if(flag)
            printf("0\n");
    }
    return 0;
}