Seek the Name, Seek the Fame
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 16036 | Accepted: 8159 |
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
此题意思是让输出字符串中所有具有相同前后缀时的字符串长度。
已知 next数组中存放的是当前字符之前的字符串中具有相同前后缀的长度，
例如：next[i]=5;表示从字符串开始到下标为 i-1时，具有长度为5的相同前后缀子串
字符串：ababcababababcabab
下标： 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
字符： a b a b c a b a b a b a b c a b a b
next： -1 0 0 1 2 0 1 2 3 4 3 4 3 4 5 6 7 8 9
1、长度为 18 时，字符串为 ababcababababcabab，前缀 ababcabab，后缀 ababcabab，输出18；
2、长度为 9 时，字符串为 ababcabab，前缀 abab，后缀abab，输出 9；
3、长度为 4 时，字符串为 abab，前缀 ab，后缀 ab，输出 4；
4、长度为 2 时，字符串为 ab，前缀 a，后缀 b，此时前后缀不相同，输出此时的字符串长度 2，并结束；
（因为字符串是从后向前一一查找判断的，所以当找到不想同的前后缀时既查找完毕，输出并结束）
......
如果还不明白，我就仍然以上述例子说，next[18]=9;意思是说，此字符串有长度为 9 的相同前后缀， 如上 1 所示，
设字符串长度为 9，此时next[9]=4,表示在此长度为 9 的前缀中，又有长度为 4 的相同的前后缀，
设字符串长度为 4，此时next[4]=2;.............
依次进行....
具体代码：
#include <stdio.h>
#include <string.h>
char s[400005];
int next[400005];
int ans[400005];
void Make_next(int len)
{
int i=0,j=-1;
memset(next,0,sizeof(next));
next[0]=-1;
while(i<len)
{
if(j==-1 || s[i]==s[j])
{
i++; j++;
next[i]=j;
}
else
j=next[j];
}
}
int main()
{
while(scanf("%s",&s)!=EOF)
{
int len=strlen(s);
Make_next(len);
int i,j=0;
memset(ans,0,sizeof(ans));
for(i=len;next[i]!=-1;)//直到找到没有前缀后缀相同为止
{
ans[j++]=i;//记录具有相同前后缀时的字符串的长度
i=next[i];
}
for(i=j-1;i>=0;i--)
printf("%d ",ans[i]);
printf("\n");
}
return 0;
}