HDU 2594 Simpsons’ Hidden Talents

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3262    Accepted Submission(s): 1229

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：求第一个字符串的开头和第二个字符串结尾相同子串的长度，如果不相同，输出0

思路：KMP模板简单运用，判断两串匹配后返回值是否为零，如果不为零就输出相应长度，否则输出0。还是KMP题目做的太少了，总觉得理解起来还有些吃力。。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define N 100009
using namespace std;

char a[N],b[N];
int next[N];
int en[N];
int len1;
int len2;

void getnext()
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;

    while(i<len1)
    {
        if(j==-1||a[i]==a[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
        j=next[j];
    }
}

void getend()
{
    getnext();

    int i,j;
    i=0;
    j=0;
    en[0]=0;

    while(i<len2)
    {
        if(j==-1||a[j]==b[i])
        {
            i++;
            j++;
            en[i]=j;
        }
        else

        j=next[j];
    }
}

int main()
{

while(~scanf("%s%s",&a,&b))
{
       len1=strlen(a);
       len2=strlen(b);

       getend();

       //for(int i=0;i<=len1;i++)
      // cout<<next[i]<<" ";
      // cout<<endl;

       //for(int i=0;i<=len2;i++)
       //cout<<en[i]<<" ";
      // cout<<endl;

       if(en[len2]==0) puts("0");

       else
       {
           for(int i=0;i<en[len2];i++)
           cout<<a[i];

           cout<<" "<<en[len2]<<endl;
       }

}
    return 0;
}

<pre name="code" class="cpp">#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>

using namespace std;

char s1[55555],s2[55555];

int nt[55555];
int en[55555];

void getnext(int len1)
{
    int k=-1;//此处一定要初始化为-1
    int j=0;
    nt[0]=-1;

    while(j<len1)
    {
        if(k<0 || s1[j]==s1[k])
        {
            k++;
            j++;
            nt[j]=k;
        }
        else
            k=nt[k];
    }
}

void getend(int len2)
{
    int k=0;//初始化为0，否则会wa。因为可能两个串完全相同
    int j=0;
    en[0]=-1;
    while(j<len2)
    {
        if(k<0 || s1[k]==s2[j])//类似于自身匹配的一个互相匹配过程
        {
            k++;
            j++;
            en[j]=k;
        }
        else
            k=nt[k];
    }
}

int main()
{
    int len1,len2;
    while(~scanf("%s",s1))
    {
        scanf("%s",s2);
        len1=strlen(s1);
        len2=strlen(s2);

        memset(en,0,sizeof en);
        memset(nt,0,sizeof nt);

        getnext(len1);
        getend(len2);
//
//        for(int i=0;i<=len1;i++)
//            cout<<nt[i]<<" ";
//            cout<<endl;

//
//        for(int i=0;i<=len2;i++)
//            cout<<en[i]<<" ";
//        cout<<endl;

        if(en[len2]==0)
            printf("%d\n",0);
        else
        {
            for(int i=0;i<en[len2];i++)
                cout<<s1[i];
            cout<<" "<<en[len2]<<endl;
        }

    }

    return 0;
}