hdu 4468 spy(KMP应用，5级）

Spy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 285    Accepted Submission(s): 136

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 10⁵). You may assume that the input contains no more than 2 × 10⁶ 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

abcaababcadabcabcadaaabbbaaaabbbaaabcababcd

 

Sample Output

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

 

Source

2012 Asia Chengdu Regional Contest

 

Recommend

liuyiding

 

思路：每次正常kmp匹配，当能覆盖到的就不加，不能覆盖到的，就把那一段加到s中.

#include<cstdio>#include<iostream>#include<cstring>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))using namespace std;const int mm=1e5+9;char s[mm],t[mm];int f[mm];int main(){ int len,pos,last,cas=0;  while(~scanf("%s",s+1))  {    len=strlen(s+1);    pos=1;clr(t,0);    t[1]=s[1];    int j=0;last=0;    FOR(i,1,len)    {      while(j&&s[i]!=t[j+1])j=f[j];///正常匹配      if(s[i]==t[j+1])++j;      if(!j)///都没有可匹配的。就加上这一段      {        int p=f[pos];        FOR(k,last+1,i)        {          t[++pos]=s[k];          while(p&&t[pos]!=t[p+1])p=f[p];///找匹配          if(t[pos]==t[p+1])f[pos]=p+1;          else f[pos]=0;        }        j=pos;      }      if(j==pos)last=i;    }    printf("Case %d: %d\n", ++cas,len+pos-last);  }}