POJ—2752 Seek the Name, Seek the Fame

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本文深入探讨了一种用于解决字符串匹配问题的创新算法，通过实例详细解释了如何利用该算法找到字符串中的前缀-后缀子串，从而为解决实际问题提供了一种高效的方法。文章通过具体的代码实现，展示了算法的工作原理和步骤。

Time Limit: 2000MSMemory Limit: 65536KTotal Submissions: 24952Accepted: 12999DescriptionThe little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:

Step1. Connect the father’s name and the mother’s name, to a new string S.

Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).

Example: Father=‘ala’, Mother=‘la’, we have S = ‘ala’+‘la’ = ‘alala’. Potential prefix-suffix strings of S are {‘a’, ‘ala’, ‘alala’}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)

InputThe input contains a number of test cases. Each test case occupies a single line that contains the string S described above.

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.

OutputFor each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby’s name.Sample Inputababcababababcabab
aaaaa
Sample Output2 4 9 18
1 2 3 4 5
SourcePOJ Monthly–2006.01.22,Zeyuan Zhu

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
int next[400005];
void Next(char T[])
{
 int l=strlen(T),i=0,k=-1;
 next[0]=-1;
 while(i<l)
 {
  if(T[i]==T[k]||k==-1)next[++i]=++k;
  else k=next[k];
 }
}
int main()
{
 char T[400005];
 while(scanf("%s",&T)!=EOF)
 {
  stack<int>s;
  int i=strlen(T);
  Next(T);
  while(i>0)
  {
   s.push(i);
   i=next[i];
  }
  while(!s.empty())
  {
   printf("%d",s.top());
   s.pop();
   if(!s.empty())printf(" ");
   else printf("\n");
  }
 }
 return 0;
}