标签：without example open init uitable determine nothing 理解 div
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string s, carved on a rock below the temple‘s gates. Asterix supposed that that‘s the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring t of the string s.
Prefix supposed that the substring t is the beginning of the string s; Suffix supposed that the substring t should be the end of the string s; and Obelix supposed that t should be located somewhere inside the string s, that is, t is neither its beginning, nor its end.
Asterix chose the substring t so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring t aloud, the temple doors opened.
You know the string s. Find the substring t or determine that such substring does not exist and all that‘s been written above is just a nice legend.
You are given the string s whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
Print the string t. If a suitable t string does not exist, then print "Just a legend" without the quotes.
fixprefixsuffix
fix
abcdabc
Just a legend
完全没有想法。先开始脑抽了，想了个二分，很激动的认为对了，结果wa，然后发现公共前缀的公共后缀的长度似乎不能二分。。。gg
又学了一下kmp。。。之前学的忘光了。。。
还是不太懂（mark）
这道题巧妙地利用了kmp中的nxt，这里的nxt就是kmp中预处理的失配函数。nxt[i]表示的是[1-i]位置的字符串中最长的前缀和后缀的公共部分的长度(s=abcab的nxt[5]=2,因为s[1-2]=s[4-5])
我们要求的正好和nxt所表示的东西很相似，于是就用上。
求完nxt，把每个长度标记。nxt[n]不标记，因为这是开头和结尾，不能算作中间的最长长度
从nxt[n](n是字符串长度)开始,表示的是整个串的前后缀最长公共部分的长度，如果这个长度存在，那么就可以了。（因为这个是最大的nxt,nxt是递减的）
挖个坑，似乎还是不是很理解
#include<cstdio> #include<cstring> using namespace std; #define N 1000010 int n; int nxt[N],used[N]; char s[N]; void Init() { int j=0; for(int i=2;i<=n;i++) { while(j&&s[i]!=s[j+1]) j=nxt[j]; if(s[i]==s[j+1]) j++; nxt[i]=j; } } int main() { scanf("%s",s+1); n=strlen(s+1); Init(); for(int i=1;i<n;i++) used[nxt[i]]=1; used[0]=0; for(int i=n;i;i=nxt[i]) if(used[nxt[i]]) { for(int j=1;j<=nxt[i];j++) printf("%c",s[j]); return 0; } printf("Just a legend"); return 0; }
标签：without example open init uitable determine nothing 理解 div
原文地址：http://www.cnblogs.com/19992147orz/p/6284076.html
