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

 

题意：给定两个串A，B，求最长的A的前缀S使得S也是B的后缀。

可以把B直接拼接到A的后面，为了避免前缀长度大于A可以先在A的后面加上一个不可能

匹配的字符，然后通过next数组求出拼接串的最长前缀（等于后缀）。

#include 
#include 
#include 
#include 
#include 
#include 
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]) {//匹配  
            j++, t++;  
            next[j] = t;
        }  
        else //失配  
            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\n", next[n]);
    }
    return 0;
}