HDU 2594 (KMP)

Simpsons’ Hidden Talents
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5819    Accepted Submission(s): 2101
Problem 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
   
   
   
   

 
題意:2つの列A,Bを与えて、最も長いAの接頭辞Sを求めてSもBの接尾辞になるようにする.
Bを直接Aの後ろにつなぎ合わせることができて、接頭辞の長さがAより大きいことを避けるために先にAの後ろに1つの不可能を加えることができます
一致する文字は、next配列によって接合列の最長接頭辞(接尾辞に等しい)を求めます.

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
#define maxn 111111

char T[maxn], P[maxn];
int n, m;
#define next Next
int next[maxn];

void get_next (char *p) {  
    int m = strlen (p);
    int t;  
    t = next[0] = -1;  
    int j = 0;  
    while (j < m) {  
        if (t < 0 || p[j] == p[t]) {//    
            j++, t++;  
            next[j] = t;
        }  
        else //    
            t = next[t];  
    } 
}  

int main () {
    while (scanf ("%s%s", T, P) == 2) { 
        n = strlen (T);
        m = strlen (P);
        T[n] = '*';
        for (int j = 0; j < m; j++) 
            T[j+1+n] = P[j];
        n = m+n+1;
        T[n] = '\0';
        get_next (T);
        if (next[n]) {
            for (int i = 0; i < next[n]; i++) printf ("%c", T[i]);
            printf (" ");
        }
        printf ("%d
", next[n]);
    }
    return 0;
}