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  • KMP 专场 POJ2752

    F - kmp
    Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
    Submit

    Status

    Practice

    POJ 2752
    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

    /*********************************
         author   : Grant Yuan
         algorithm; kmp
         source   : POJ 2752
    	  time    ;2014/10/3 20:38
     *********************************/   
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #define MAX  400007
    
    using namespace std;
    
    int next[MAX];
    char s[MAX];
    int ans;
    int l;
    int sum[MAX];
    
    void get_next()
    {
        next[0]=-1;
        int j=-1;
        int i=0;
        while(i<l){
           if(j==-1||s[j]==s[i])
           {
               i++;j++;
               next[i]=j;
           }
           else j=next[j];
        }
    }
    
    int main()
    {
        while(~scanf("%s",s)){
    		int j;
    		memset(sum,0,sizeof(sum));
            l=strlen(s);
            get_next();
            int k=0;
            for(int i=l;i!=0;)
            {
                sum[k++]=next[i];
               i=next[i];
            }
            for(int i=k-2;i>=0;--i)
                 printf("%d ",sum[i]);
           printf("%d
    ",l);}
    
        return 0;
    }


     

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  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254422.html
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