zoukankan      html  css  js  c++  java
  • poj2752 Seek the Name, Seek the Fame(next数组的运用)

    题目链接:http://poj.org/problem?id=2752


    Seek the Name, Seek the Fame
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 12018   Accepted: 5906

    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
    

    Source


    题意:寻找前子串与后子串相等的子串。

    代码例如以下:


    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    using namespace std;
    #define MAXN 1000017
    int next[MAXN];
    int ans[MAXN];
    int len;
    void getnext( char T[])
    {
        int i = 0, j = -1;
        next[0] = -1;
        while(i < len)
        {
            if(j == -1 || T[i] == T[j])
            {
                i++,j++;
                next[i] = j;
            }
            else
                j = next[j];
        }
    }
    
    int main()
    {
        char ss[MAXN];
        int length;
        while(~scanf("%s",ss))
        {
            len = strlen(ss);
            getnext(ss);
    		int n = 0 ;int i = len;
    		ans[0]=len;
    		while(next[i] > 0 )//倒着扫描next数组
    		{//递归查找前子串和后子串相等的子串
    			i = next[i];
    			ans[++n] = i ;
    		}
    		for( i = n ; i >= 0 ; i--)
    			printf("%d ",ans[i]);
    		printf("
    ");
        }
        return 0;
    }
    


  • 相关阅读:
    一些基本的操作,编译,构建,单元测试,安装,网站生成和基于Maven部署项目。
    Maven目标
    Maven是什么?
    Maven的生命周期是为了对所有的构建过程进行了抽象了,便于统一。
    mvn archetype:generate 创建Maven项目
    Maven是一个项目管理工具
    Maven项目对象模型(POM)
    e816. 创建工具栏
    e836. 设置JTabbedPane中卡片的提示语
    e834. 设置JTabbedPane中卡片的位置
  • 原文地址:https://www.cnblogs.com/bhlsheji/p/5085627.html
Copyright © 2011-2022 走看看