poj 4468Spy(kmp算法)

解密信息：利用KMP算法解决字符串匹配问题

最新推荐文章于 2025-09-12 18:08:49 发布

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最新推荐文章于 2025-09-12 18:08:49 发布

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文章标签： java

原文链接：http://www.cnblogs.com/xinyuyuanm/p/3212005.html

本文详细介绍了如何通过KMP算法解决字符串匹配问题，包括定义串B由串A的前缀组成并附加A本身，以及如何求解给定串B时的最小串A长度。通过实例和代码实现，深入理解KMP算法的应用。

Spy

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 204 Accepted Submission(s): 96

Problem Description

“Be subtle! Be subtle! And use your spies for every kind of business. ”
— Sun Tzu
“A spy with insufficient ability really sucks”
— An anonymous general who lost the war
You, a general, following Sun Tzu’s instruction, make heavy use of spies and agents to gain information secretly in order to win the war (and return home to get married, what a flag you set up). However, the so-called “secret message” brought back by your spy, is in fact encrypted, forcing yourself into making deep study of message encryption employed by your enemy.
Finally you found how your enemy encrypts message. The original message, namely s, consists of lowercase Latin alphabets. Then the following steps would be taken:
* Step 1: Let r = s
* Step 2: Remove r’s suffix (may be empty) whose length is less than length of s and append s to r. More precisely, firstly donate r[1...n], s[1...m], then an integer i is chosen, satisfying i ≤ n, n - i < m, and we make our new r = r[1...i] + s[1...m]. This step might be taken for several times or not be taken at all.
What your spy brought back is the encrypted message r, you should solve for the minimal possible length of s (which is enough for your tactical actions).

Input

There are several test cases.
For each test case there is a single line containing only one string r (The length of r does not exceed 10 ⁵). You may assume that the input contains no more than 2 × 10 ⁶ characters.
Input is terminated by EOF.

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is the desired answer.

Sample Input

    abc aab abcadabcabcad aaabbbaaaabbbaa abcababcd 
  

Sample Output

    Case 1: 3 Case 2: 2 Case 3: 5 Case 4: 6 Case 5: 4 
  

Source

2012 Asia Chengdu Regional Contest

第一次做这种题也是第一次接触kmp算法，感觉很糟糕，这个题有好多不懂的地方，看着解题报告写的，写完了提交78ms，看人家的好多31ms，想自己优化一下，改了半天还是78ms，可能还是因为没有真正理解它吧，不过以后还会继续看的，继续练习，总会学会的

说一下题意，我觉得这个题目还挺难理解的，可能是英语太差了，连着看了好多解题报告才把题目搞懂，悲催。。。定义串B是由串A的若干前缀加在一起再加上串A本身构成的。现在给出串B，求串A的最小长度。我解释一下这个若干前缀的意思，比如A串为abcdef，a、ab、abc、abcd等都可以说是A的前缀，也就是从前往后的某一段字符串

#include<stdio.h>
#include<string.h>
char s[100050],c[100050];
int next[100050],n,cn;

void match()
{
	int i,j=0,tem,k,a;
	for(i=1;i<=n;i++)
	{
		while(j&&c[j+1]!=s[i])
			j=next[j];
		if(s[i]==c[j+1])
			j++;
		if(j==0)
		{
			for(k=tem;k<=i;k++)
			{
				c[++cn]=s[k];
					a=cn;
				if(c[a]==c[next[a-1]+1])
					next[a]=next[a-1]+1;
				}
				tem=i+1;
			}
		}
		else if(j==cn)
			tem=i+1;//记录当前s位置的下一个位置
	}
	cn+=n-tem+1;
}

int main()
{
	int i,j,cas=1;
	while(scanf("%s",s+1)!=EOF)
	{
		n=strlen(s+1);
		memset(c,0,sizeof(c));
		memset(next,0,sizeof(next));
		c[1]=s[1];
		cn=1;
		match();
		printf("Case %d: %d\n",cas++,cn);
	}
	return 0;
}

转载于:https://www.cnblogs.com/xinyuyuanm/p/3212005.html