个人觉得kmp是很神奇的一种算法,最近刚刚开始入门
hdu 4763
Theme Section
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3162 Accepted Submission(s): 1486
Problem Description
It's
time for music! A lot of popular musicians are invited to join us in
the music festival. Each of them will play one of their representative
songs. To make the programs more interesting and challenging, the hosts
are going to add some constraints to the rhythm of the songs, i.e., each
song is required to have a 'theme section'. The theme section shall be
played at the beginning, the middle, and the end of each song. More
specifically, given a theme section E, the song will be in the format of
'EAEBE', where section A and section B could have arbitrary number of
notes. Note that there are 26 types of notes, denoted by lower case
letters 'a' - 'z'.
To get well prepared for the festival, the hosts want to know the maximum possible length of the theme section of each song. Can you help us?
To get well prepared for the festival, the hosts want to know the maximum possible length of the theme section of each song. Can you help us?
Input
The
integer N in the first line denotes the total number of songs in the
festival. Each of the following N lines consists of one string,
indicating the notes of the i-th (1 <= i <= N) song. The length of
the string will not exceed 10^6.
Output
There
will be N lines in the output, where the i-th line denotes the maximum
possible length of the theme section of the i-th song.
Sample Input
5
xy
abc
aaa
aaaaba
aaxoaaaaa
Sample Output
0
0
1
1
2
题目意思是一个字符串中前缀后缀相等且能在中间找到,重要的一个点是三者不能重叠
1 #include<iostream> 2 #include<string> 3 4 using namespace std; 5 6 int net[1000005]; 7 string a; 8 9 void get() 10 { 11 net[0]=-1; 12 int i = 0, j = -1; 13 while(i<a.length()) 14 { 15 if(j==-1||a[i]==a[j]) 16 { 17 i++,j++; 18 net[i]=j; 19 } 20 else 21 j = net[j]; 22 } 23 } 24 int main() 25 { 26 int t; 27 cin>>t; 28 while(t--) 29 { 30 cin>>a; 31 int sum = 0; 32 for(int i = 1; i < a.length(); i++) 33 if(a[0]!=a[i]) 34 sum=1; 35 if(!sum) 36 sum=a.length()/3; 37 else 38 { 39 sum=0; 40 get(); 41 int ans = net[a.length()]; 42 for(int i = ans; i >0 && !sum; i=net[i]) 43 { 44 for(int j = i*2; j <= a.length()-i; j++) 45 { 46 if(net[j]>=i) 47 sum=i; 48 } 49 } 50 } 51 cout<<sum<<endl; 52 } 53 return 0; 54 }