POJ2752 Seek the Name, Seek the Fame KMP(KMP中对next数组原理的理解)

最新推荐文章于 2022-08-23 22:16:41 发布

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这篇博客介绍了如何使用KMP算法来寻找一个由父母名字连接成的字符串的前缀后缀，并给出了限制条件和样例输入输出。通过理解KMP算法中的next数组，可以找到所有可能的前缀后缀长度。

Problem 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

思路：

KMP中对next数组原理的理解

代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN=1000000+100;
int ans[MAXN];
char P[MAXN];
int f[MAXN];
int m;
void getFail(char *P,int *f)
{
    f[0]=f[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=f[i];
        while(j && P[i]!=P[j]) j=f[j];
        f[i+1]=(P[i]==P[j])?j+1:0;
    }
}
int main()
{
    int count;
    while(scanf("%s",P)!=EOF)
    {
        count=0;
        m=strlen(P);
        getFail(P,f);
        while(m!=0)
        {
            ans[count++]=m;
            m=f[m];
        }
        for(int i=count-1;i>=0;i--)
        {
            printf("%d ",ans[i]);
        }
        printf("\n");
    }
    return 0;
}