hust 1010 The Minimum Length (KMP)

最新推荐文章于 2018-12-08 21:39:43 发布

原创最新推荐文章于 2018-12-08 21:39:43 发布 · 308 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#kmp #hust

KMP 专栏收录该内容

12 篇文章

订阅专栏

本文介绍了一种算法，用于解决给定由一个未知最短子串重复构成的新字符串时，如何找到并输出该最短子串的实际长度。通过构建Next数组来优化匹配过程，实现了高效地解决此问题。

1010 - The Minimum Length

时间限制：1秒内存限制：128兆

3105 次提交 1148 次通过

题目描述

There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA...... Now I cut it from two different position and get a new string B. Then, give you the string B, can you tell me the length of the shortest possible string A. For example, A="abcdefg". I got abcdefgabcdefgabcdefgabcdefg.... Then I cut the red part: efgabcdefgabcde as string B. From B, you should find out the shortest A.

输入

Multiply Test Cases. For each line there is a string B which contains only lowercase and uppercase charactors. The length of B is no more than 1,000,000.

输出

For each line, output an integer, as described above.

样例输入

bcabcab
efgabcdefgabcde

样例输出

3
7

分析：由于这个串是全部由一个子串连成的，所以用主串的长度减去最后一个的next值就是子串的长度。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX_N 1000005
using namespace std;
char p[MAX_N];
int ne[MAX_N];
void makeNext(const char p[],int ne[])
{
    int len=strlen(p);
    ne[0]=0;
    for(int i=1,k=0;i<len;i++)
    {
        while(k>0&&p[i]!=p[k]) k=ne[k-1];
        if(p[i]==p[k]) k++;
        ne[i]=k;
    }
}
int main()
{
    while(~scanf("%s",p))
    {
        makeNext(p,ne);
        int len=strlen(p);
        printf("%d\n",len-ne[len-1]);
    }
    return 0;
}