POJ 2752 Seek the name,seek the fame KMP next数组的应用

 

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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到现在依旧觉得，神奇，很神奇很神奇……
原来next数组还可以这样用……Orz
next数组存储的是当前长度的子串与原串前缀与子串后缀相同字符的个数，next数组本身就已经包含答案了
感觉这个讲的很清楚，不懂得可以参考：http://972169909-qq-com.iteye.com/blog/1071548

#include<iostream>  
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
int n,lb,la,crt;
char a[400005];
int nextt[400005];
int res[400005];
void kmp()
{
    int k = -1, j = 0;
    nextt[0] = -1;
    while (j < la)
    {
        if (k == -1 || a[j] == a[k])
        {
            j++; k++;
            nextt[j] = k;
        }
        else k = nextt[k];
    }
}
void solve(int n)               //递归写法参考大佬的，感觉简介又清晰……
{
    if (nextt[n] > 0)
    {
        solve(nextt[n]);
        printf("%d ", nextt[n]);
    }
}
int main()
{
    while (scanf("%s", a)!=EOF)
    {
        la = strlen(a);
        kmp();
        solve(la);
        printf("%d\n", la);
    }
    return 0;
}