poj 2752 Seek the Name, Seek the Fame

最新推荐文章于 2021-12-13 14:38:22 发布

转载最新推荐文章于 2021-12-13 14:38:22 发布 · 60 阅读

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原文链接：http://www.cnblogs.com/TheRoadToTheGold/p/6484416.html

本文介绍了一个KMP算法的实际应用案例，旨在解决一个有趣的字符串匹配问题：寻找字符串中既是前缀也是后缀的所有子串长度。文章通过一个具体的编程示例详细讲解了如何使用KMP算法来高效地解决这个问题。

Seek the Name, Seek the Fame

http://poj.org/problem?id=2752

Time Limit: 2000MS		Memory Limit: 65536K

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
题目大意：求前k字符=后k个字符 所有满足条件的k

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int t,n,len,dp[400001];
char s[400001];
int f[400010];
int ans[400010];
void getnext()
{
    for(int i=1;i<len;i++)
    {
        int j=f[i];
        while(j&&s[i]!=s[j]) j=f[j];
        f[i+1]= s[i]==s[j] ? j+1:0;
    }
}
int main()
{
    while(cin>>s)
    {
        len=strlen(s);
        getnext();
        int p=f[len];
        int t=0;
        ans[++t]=len;
        while(p)
        {
            ans[++t]=p;
            p=f[p];
        }
        for(int i=t;i;i--) printf("%d ",ans[i]);
        printf("\n");    
    }
}

做了5道kmp了，第一次没看题解自己做出来1道，感觉好失败

转载于:https://www.cnblogs.com/TheRoadToTheGold/p/6484416.html