POJ 2752 KMP算法中next数组的应用

最新推荐文章于 2022-04-06 20:14:30 发布

iteye_20746

最新推荐文章于 2022-04-06 20:14:30 发布

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本文介绍了一个利用Next数组解决字符串匹配问题的方法，并通过一个具体的编程挑战——为新生儿取名来展示Next数组的应用。文章提供了一段AC代码作为实现示例。

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这道题是昨天晚上9点多看的，今天早上来了后想了想，，感觉有思路了，吃过饭后就ac了，，，没注意到数据大小，，吃了一次re。。。。。。。这道题算是next数组的一个应用，，，有点折半查找的感觉，，，求字符串循环节和这道题，都是next数组的妙用，，，看来对next数组有一个清晰的理解非常必要啊！题目：

Seek the Name, Seek the Fame

Time Limit:2000MS		Memory Limit:65536K
Total Submissions:6869		Accepted:3207

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

ac代码：

#include <iostream> #include <cstdio> #include <string.h> const int N=400010; char str[N]; int nextt[N],num[N]; void get_nextt(int len){ int i,j=0; nextt[0]=0; for(i=1;i<len;++i){ int temp=nextt[i-1]; while(temp&&str[i]!=str[temp]) temp=nextt[temp-1]; if(str[i]==str[temp]) nextt[i]=temp+1; else nextt[i]=0; } } int main(){ while(scanf("%s",str)!=EOF){ int len=strlen(str); get_nextt(len); int j=1; for(int i=nextt[len-1];i>=1;i=nextt[i-1]){ num[j++]=i; } num[0]=len; for(int i=j-1;i>=0;i--) printf("%d ",num[i]); printf("\n"); } return 0; }