HUST 1010 (KMP 水~)

最新推荐文章于 2019-09-27 08:51:57 发布

morejarphone

最新推荐文章于 2019-09-27 08:51:57 发布

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本文探讨如何从无限循环字符串中找到最短原始字符串，涉及字符串操作与模式匹配算法的应用。

1010 - The Minimum Length

时间限制：1秒内存限制：128兆

1693 次提交 623 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

题意：有一个串A，用A作为循环节构造无穷长的串C：AAAAAA....，然后截取中间的一

段串B，求最短的A。

容易发现B中的最小循环节长度就是A的最短长度，于是直接n-next[n]。

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
#define maxn 1111111

char T[maxn];
int n;
#define next Next
int next[maxn];

void get_next (char *p) {  
    int m = strlen (p);
    int t;  
    t = next[0] = -1;  
    int j = 0;  
    while (j < m) {  
        if (t < 0 || p[j] == p[t]) {//匹配  
            j++, t++;  
            next[j] = t;
        }  
        else //失配  
            t = next[t];  
    } 
}  

int main () {
    while (scanf ("%s", T) == 1) {
        get_next (T); 
        n = strlen (T);
        printf ("%d\n", n-next[n]);
    }
    return 0;
}