Simpsons’ Hidden Talents(kmp next数组的应用)

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7565    Accepted Submission(s): 2688

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

 

Source

HDU 2010-05 Programming Contest

把第一个字符串和第二个字符串连在一起，然后用kmp种求next数组的模板，找到这两个字符串种符合题目要求的字串。注意如果第一个字符串的长度为len1，第二个字符串的长度为len2。我们不能直接把next[len1+len2]作为结果，因为，next[len1+len2]的长度有可能大于第一个或第二个字符串。具体操作看代码。

#include<iostream>#include<string.h>using namespace std;string a,b;int _next[100005];void Get_Next(string str){      _next[0]=-1;      int j=-1,i=0;      while(i<str.length()){          if(j==-1||str[j]==str[i]){              j++;i++;              _next[i]=j;          }          else j=_next[j];      }  }  int main(){while(cin>>a>>b){string c=a+b;Get_Next(c);int len=a.length()+b.length();int k=_next[len];while(k>a.length()||k>b.length()){k=_next[k];}//cout<<next[k]<<endl;if(k==0){cout<<0<<endl;}else{for(int i=0;i<k;i++){cout<<c[i];}cout<<" "<<k<<endl;}}}