POJ 2752 Seek the Name, Seek the Fame（KMP+next数组的运用）

拼搏现实的明天。 2022-06-10 14:08 126阅读 0赞

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

题解：

题意：

给你一个串，让你求该串的前缀和后缀相同相同处的长度，按从小到大排列，next数组里面就是前缀和后缀能匹配到的长度，直接用next数组还原一下就好了

代码：

#include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<vector>
    #include<deque>
    using namespace std;
    #define lson k*2
    #define rson k*2+1
    #define M (t[k].l+t[k].r)/2
    #define INF 1008611111
    #define ll long long
    #define eps 1e-15
    int Next[400005];
    char T[400005];
    int a[400005];
    void init()
    {
        int i=0,j=-1;
        Next[0]=-1;
        while(T[i])
        {
            if(j==-1||T[i]==T[j])
            {
                i++;
                j++;
                Next[i]=j;
            }
            else
                j=Next[j];
        }
    }
    int main()
    {
        int i,j,n,len,cent;
        while(gets(T)!=NULL)
        {
            init();
            len=strlen(T);
            cent=0;
            i=len;
            while(i!=0)
            {
                a[cent]=Next[i];
                cent++;
                i=Next[i];
            }
            for(i=cent-2;i>=0;i--)
                printf("%d ",a[i]);
            printf("%d\n",len);
        }
        return 0;
    }