Palindrome
Time Limit: 15000MS
Memory Limit: 65536K
Total Submissions: 12214
Accepted: 4583
Description
Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the length of the largest palindrome in a string?"
A string is said to be a palindrome if it reads the same both forwards and backwards, for example "madam" is a palindrome while "acm" is not.
The students recognized that this is a classical problem but couldn't come up with a solution better than iterating over all substrings and checking whether they are palindrome or not, obviously this algorithm is not efficient at all, after a while Andy raised his hand and said "Okay, I've a better algorithm" and before he starts to explain his idea he stopped for a moment and then said "Well, I've an even better algorithm!".
If you think you know Andy's final solution then prove it! Given a string of at most 1000000 characters find and print the length of the largest palindrome inside this string.
Input
Your program will be tested on at most 30 test cases, each test case is given as a string of at most 1000000 lowercase characters on a line by itself. The input is terminated by a line that starts with the string "END" (quotes for clarity).
Output
For each test case in the input print the test case number and the length of the largest palindrome.
Sample Input
abcbabcbabcba
abacacbaaaab
END
Sample Output
Case 1: 13
Case 2: 6
题解
求解最长回文子串的长度
本来是想练hash的,思路:hash+二分,时间复杂度(O(T*len*log(len))),其中T为组数,最大为30,len最大为1e6,肯定是过不了的,但是抱着既然是练习hash的心请,多打几遍也没事.然后就硬着头皮写下去,结果一个上午没有调出来,天衣无缝~~~
然后我含恨而去,学习了manacher,发现挺简单的,想学习的推荐博客super_hyper(真的很好理解,讲得很好)
原代码我一定会调出来的
Code
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define Min(a,b) (a)<(b)?(a):(b)
#define Max(a,b) (a)>(b)?(a):(b)
#define in(i) (i=read())
using namespace std;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') {ans=(ans<<1)+(ans<<3)+i-'0'; i=getchar();}
return ans*f;
}
int n;
char s[1000010],s_new[2000010];
int p[2000010];
int init() {
int len=strlen(s),j=2;
s_new[0]='$'; s_new[1]='#';
for(int i=0;i<len;i++) {
s_new[j++]=s[i];
s_new[j++]='#';
}
s_new[j]=' ';
return j;
}
int Manacher() {
int len=init();
int max_len=-1,id,mx=0;
for(int i=1;i<len;i++) {
if(i<mx) p[i]=Min(p[2*id-i],mx-i);
else p[i]=1;
while(s_new[i-p[i]]==s_new[i+p[i]]) p[i]++;
if(mx<i+p[i]) {
id=i; mx=i+p[i];
}
max_len=Max(max_len,p[i]-1);
}
return max_len;
}
int main()
{
int cnt=0;
while(1) {
cnt++;
scanf("%s",s);
if(s[0]=='E' && s[1]=='N' && s[2]=='D') break;
printf("Case %d: %d
",cnt,Manacher());
memset(s,0,sizeof(s));
}
return 0;
}