kuangbin KMP H题

最新推荐文章于 2020-11-27 20:26:56 发布

西瓜君wtml

最新推荐文章于 2020-11-27 20:26:56 发布

阅读量386

点赞数

分类专栏： KMP

本文链接：https://blog.youkuaiyun.com/q451792269/article/details/77969492

版权

KMP 专栏收录该内容

17 篇文章

订阅专栏

本文介绍了一种用于寻找字符串中前后缀匹配部分的高效算法。通过使用next数组来优化搜索过程，该算法能快速找到所有可能的前后缀匹配，并输出它们的长度。适用于处理大量字符串数据的应用场景。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

H - Seek the Name, Seek the Fame

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：

next数组的妙用。
仔细观察。
最小的前后缀一定在比他长的前后缀中出现过。
我们可以利用这一性质，[0,next[i]]出现过一定在[0,i]出现过。
141ms

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

const int MAXN = 400000 + 10;

char str[MAXN];
int len;
int nt[MAXN];

void getNext()
{
    nt[0] = -1;
    int i = 0, j = -1;
    while (i <= len)
    {
        if (j == -1 || str[i] == str[j])
        {
            nt[++i] = ++j;
        }
        else
        {
            j = nt[j];
        }
    }
}
vector<int> ans;
int main()
{
    while(scanf("%s",str)!=EOF)
    {
        len = strlen(str);
        getNext();

        while(len>0)
        {
            ans.push_back(len);
            len=nt[len];
        }
        int siz = ans.size();

        for(int i=siz-1;i>=1;i--)
        {
            printf("%d ",ans[i]);
        }
        printf("%d\n",ans[0]);
        ans.clear();

    }
    return 0;
}