HDU 2594 Simpsons’ Hidden Talents(KMP理解應用)

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1658    Accepted Submission(s): 605


Problem DescriptionHomer: 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. 

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

OutputOutput 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

clintonhomerriemannmarjorie

 

Sample Output

0rie 3

 

SourceHDU 2010-05 Programming Contest 

Recommendlcy 首先想說的是這個題目和上一篇一樣不要要注意的是求next 的next時候直到next比兩個串長度都小的時候為止

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;char str[110000];int next[110000];int get_next(){    int i=0,j=-1;    next[0]=-1;    while(str[i])    {        if(j==-1 || str[i]==str[j])        {            i++;            j++;            next[i]=j;        }        else        j=next[j];    }    return 0;}int main(){    int len1,len2,i,j,k,len;    while(scanf("%s",str)!=EOF)    {        len1=strlen(str);        scanf("%s",str+len1);        len2=strlen(str+len1);        get_next();        len=len1+len2;        while(1)        {            if(next[len] <= len1 && next[len] <= len2)            break;            else            len=next[len];        }        if(next[len]==0 || next[len]==-1)        printf("0\n");        else        {            for(i=0;i<next[len];i++)            printf("%c",str[i]);            printf(" %d",next[len]);            printf("\n");        }    }    return 0;}