[POJ 2752] Seek the Name, Seek the Fame【KMP】

  • 2018-02-20
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Problem:

Time Limit: 2000MS Memory Limit: 65536K

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

Solution:

这又是一道考查 KMPnxt[] 数组性质的题。。

设原串 S 的长度为 len,则我们发现同时为前后缀的最长串就是 S1..nxt[len] == Slen-nxt[len]+1..len

然后我们发现,S1..nxt[nxt[len]] == Snxt[len]-nxt[nxt[len]]+1..nxt[len] == Slen-nxt[nxt[len]]+1..len 是第二长的串。

以此类推,将变量 j 从 len 处开始,每次向左跳 j = nxt[j],记录下所有 j,反序输出即可。

Code: O(T|S|) [3140K, 438MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

char S[400005];
int len, nxt[400005];
int stk[400005], tops;

int main(){
	while(scanf("%s", S + 1) != EOF){
		len = strlen(S + 1);
		nxt[1] = 0;
		for(register int i = 2, j = 0; i <= len; i++){
			while(j && S[j + 1] != S[i]) j = nxt[j];
			if(S[j + 1] == S[i]) j++;
			nxt[i] = j;
		}
		for(register int j = len; j; j = nxt[j]) stk[tops++] = j;
		while(tops > 1) printf("%d ", stk[--tops]);
		printf("%d\n", stk[--tops]);
	}
	return 0;
}

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