hdu2594（kmp）

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5429    Accepted Submission(s): 1965

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

clintonhomerriemannmarjorie

 

Sample Output

0rie 3

 

这个题是求两个字符串前一个的前缀和后一个的后缀最大公共长度呢，我的想法是将两个串连在一起，通过next数组性质求解，然而最开始忘记了这个长度一定小于两个子串最短的那个，导致wa了几次，后来改正了过来

#include <iostream>#include <stdio.h>#include <string.h>#define mem(a) memset((a),0,sizeof(a))using namespace std;char s1[50005];char s2[50005];char s[100020];int ntr[100020];void get_ntr(char *s){    int i=0,j=-1;    ntr[i]=-1;    int len=strlen(s);    while(i<len)    {        if(j==-1||s[i]==s[j])        {            i++;            j++;            ntr[i]=j;        }        else            j=ntr[j];    }}int main(){    while(~scanf("%s %s",s1,s2))    {        mem(ntr);        strcpy(s,s1);        strcat(s,s2);        get_ntr(s);        int len=strlen(s);        int len1=strlen(s1),len2=strlen(s2),flag=0,mx=-2;        if(len1<len2)        {            flag=1;            mx=len2;        }        else        {            flag=2;            mx=len1;        }        if(ntr[len]!=0)        {            if(ntr[len]<mx)            {                for(int i=0; i<ntr[len]; i++)                {                    printf("%c",s[i]);                }                printf(" %d\n",ntr[len]);            }            else            {                if(flag==1) printf("%s %d\n",s1,len1);                else printf("%s %d\n",s2,len2);            }        }        else            printf("0\n");            mem(s1);            mem(s2);    }    return 0;}