POJ 2752 Seek the Name, Seek the Fame(KMP 理解应用)

Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9735   Accepted: 4670 Description The 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:)  Input The 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.  Output For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name. Sample Input ababcababababcabab aaaaa Sample Output 2 4 9 18 1 2 3 4 5 Source POJ Monthly--2006.01.22,Zeyuan Zhu

这个题目只要对KMP有一定理解就能做了，题目意思是求一个串的所有前缀与后缀相同的串的长度

这个题目看见首先这个串就是一个，而且是最长的

然后就是对这个串求一次next求完之后就是输出结果

结果就是输出next[len]这个肯定不是本身最长的，然后不断next next就OK了

至于为什么这样其实理解了KMP是不难发现的

就拿题目中的串来讲解一下吧！

ababcababababcabab

0: -1 1: 0 2: 0 3: 1 4: 2 5: 0 6: 1 7: 2 8: 3 9: 4 10: 3 11: 4 12: 3 13: 4 14: 5 15: 6 16: 7 17: 8 18: 9 

看18位置的next是9，那么这个是一个答案，因为在位置18前面有9个字符是满足题目的串

然后跳到9的这个位置，发现就是答案，其实结果已经很明显了

18位置的next是9，那么9位置的next值前面的next值个字符和18前面的几个相同，而又与开始的

几个相同，这样就是后缀与前缀相同，就是题目要求的答案了！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char str[500000];
int next[500000];
int ans[500000];
int pos;
int get_next(int i,int j)
{
next[0]=-1;
while(str[i])
{
if(j==-1 || str[i]==str[j])
{
j++;
i++;
next[i]=j;
}
else
j=next[j];
}
return i;
}
int main()
{
int i,j,k;
int len;
while(scanf("%s",str)!=EOF)
{
pos=0;
len=get_next(0,-1);
i=next[len];
while(i!=-1)
{
ans[pos++]=i;
i=next[i];
}
for(i=pos-2;i>=0;i--)
printf("%d ",ans[i]);
printf("%d\n",len);
}
return 0;
}