poj2752 KMP水题

题目：

Status
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

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思路：

题目意思就是给一个字符串然后去找它的所有的前后缀相同的串的长度，包括本身。
很容易联想到KMP算法，如果理解了KMP算法到next数组的求法就不算太难。

• 首先去预处理求next数组。
• 然后是从k=strlen(A)开始，反复让k=next[k-1].
• 过程中经历过的k就是前后缀相同的串的长度

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代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
#define N 400000
char A[N+5];
int next[N+5];
int ans[N+5];
int sum;
void get_next()
{
    next[0]=0;
    int i,k;
    int len=strlen(A);
    for(i=1,k=0;i<len;i++)
    {
        while(k>0&&A[k]!=A[i])
            k=next[k-1];
        if(A[k]==A[i])
            k++;
        next[i]=k;
    }
    for(int i=0;i<len;i++)
        cout<<next[i]<<endl;
}
int main()
{
    while(scanf("%s",A)!=EOF)
    {
        sum=0;
        get_next();
        int len=strlen(A);
        int k=len;
        while(k>0)
        {
            ans[++sum]=k;
            k=next[k-1];
        }
        printf("%d",ans[sum]);
        for(int i=sum-1;i>0;i--)
            printf(" %d",ans[i]);
        printf("\n");
    }
    return 0;
}

要是问KMP是什么鬼，可以看看这个
http://www.cnblogs.com/c-cloud/p/3224788.html

P.S. 第一次写，挑一个水题，估计还挺好玩的orz。