D - Palindrome

Time Limit:15000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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

求最长回文串的长度。

(manacher算法)

#include<cstdio>
#include<cstring>
#include<iostream>
#define m(s) memset(s,0,sizeof s);
using namespace std;
const int N=1e6+;
int l,cas,len,p[N<<];
char s[N],S[N<<];
void manacher(){
int ans=,id=,mx=-;
for(int i=;i<l;i++){
if(id+mx>i) p[i]=min(p[id*-i],id+mx-i);
while(i-p[i]->=&&i+p[i]+<=l&&S[i-p[i]-]==S[i+p[i]+]) p[i]++;
if(id+mx<i+p[i]) id=i,mx=p[i];
ans=max(ans,p[i]);
}
printf("Case %d: %d\n",++cas,ans);
}
int main(){
while(scanf("%s",s)==){
if(s[]=='E') break;
len=strlen(s);m(p);m(S);
l=-;
for(int i=;i<len;i++) S[++l]='#',S[++l]=s[i];
S[++l]='#';
manacher();
}
return ;
}

//====================================================

//hash有点慢
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef int i64;
const int N=1e6+;
int n,m,cas,ans,a[N<<];char s[N];
i64 P,pow[N<<],hash_l[N<<],hash_r[N<<];
void get_hash(){
pow[]=;hash_r[]=hash_l[m+]=;
for(int i=;i<=m;i++) pow[i]=pow[i-]*P;
for(int i=;i<=m;i++) hash_r[i]=hash_r[i-]*P+a[i];
for(int i=m;i>=;i--) hash_l[i]=hash_l[i+]*P+a[i]; }
int main(){
P=;
while(scanf("%s",s+)==){
if(s[]=='E') break;
n=strlen(s+);ans=m=;
for(int i=;i<=n;i++){
a[++m]='#';
a[++m]=s[i]-'a';
}
a[++m]='#';
get_hash();
int l,r,mid;
for(int i=;i<=m;i++){
l=;
if(i-<m-i) r=i;
else r=m-i+;
while(r-l>){
mid=l+r>>;
i64 hash_to_l=hash_r[i-]-hash_r[i-mid-]*pow[mid];
i64 hash_to_r=hash_l[i+]-hash_l[i+mid+]*pow[mid];
if(hash_to_l==hash_to_r) l=mid;
else r=mid;
}
ans=max(ans,l);
}
printf("Case %d: %d\n",++cas,ans);
}
return ;
}

POJ 3974 Palindrome的更多相关文章

  1. POJ 3974 - Palindrome - [字符串hash+二分]

    题目链接:http://poj.org/problem?id=3974 Time Limit: 15000MS Memory Limit: 65536K Description Andy the sm ...

  2. POJ 3974 Palindrome(最长回文子串)

    题目链接:http://poj.org/problem?id=3974 题意:求一给定字符串最长回文子串的长度 思路:直接套模板manacher算法 code: #include <cstdio ...

  3. ●POJ 3974 Palindrome(Manacher)

    题链: http://poj.org/problem?id=3974 题解: Manacher 求最长回文串长度. 终于会了传说中的马拉车,激动.推荐一个很棒的博客:https://www.61mon ...

  4. POJ 3974 Palindrome 字符串 Manacher算法

    http://poj.org/problem?id=3974 模板题,Manacher算法主要利用了已匹配回文串的对称性,对前面已匹配的回文串进行利用,使时间复杂度从O(n^2)变为O(n). htt ...

  5. poj 3974 Palindrome (manacher)

    Palindrome Time Limit: 15000MS   Memory Limit: 65536K Total Submissions: 12616   Accepted: 4769 Desc ...

  6. 后缀数组 POJ 3974 Palindrome && URAL 1297 Palindrome

    题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串 ...

  7. POJ 3974 Palindrome (算竞进阶习题)

    hash + 二分答案 数据范围肯定不能暴力,所以考虑哈希. 把前缀和后缀都哈希过之后,扫描一边字符串,对每个字符串二分枚举回文串长度,注意要分奇数和偶数 #include <iostream& ...

  8. POJ 3974 Palindrome | 马拉车模板

    给一个字符串,求最长回文字串有多长 #include<cstdio> #include<algorithm> #include<cstring> #define N ...

  9. POJ 1159 Palindrome(字符串变回文:LCS)

    POJ 1159 Palindrome(字符串变回文:LCS) id=1159">http://poj.org/problem? id=1159 题意: 给你一个字符串, 问你做少须要 ...

随机推荐

  1. Sizing and Capacity Planning for SharePoint 2013 - Resources

    http://blogs.msdn.com/b/sanjaynarang/archive/2013/04/06/sizing-and-capacity-planning-for-sharepoint- ...

  2. 转 asp.net中如何退出整个框架(frameset),回到登录界面

    如: <frameset rows= "74,*,0,0 " cols= "* " frameborder= "NO " border ...

  3. MyBatis入门(四)---动态SQL

    一.创建数据库表 1.1.创建表 USE `mybatis`; /*Table structure for table `user` */ DROP TABLE IF EXISTS `user`; C ...

  4. NSDate,NSNumber,NSValue

    NSDate #import <Foundation/Foundation.h> int main(int argc, const char * argv[]) { @autoreleas ...

  5. nginx安装过程,报错处理:make[1]: *** [objs/addon/src/bson.o] Error 1

    nginx安装过程中,经常会有各种错误: 具体安装步骤这里不做说明,网上一搜大把: 主要分析安装过程中遇到的问题 在make编译的时候,若报如下错误: cc1: warnings being trea ...

  6. 每日Scrum--No.7

    Yesterday:学习和设计路线的编程 Today:编写代码 Problem:.在设计查询参观路线的时候,整个逻辑特别的混乱,设想了各种树,图以及网的遍历问题,但经过多次与同学的交流以及网上的查询资 ...

  7. javascript元素绑定事件

    js元素绑定事件   想给一个元素绑定一个方法之后,在绑定一个方法而且不被覆盖 window.onload = function () { alert('a'); } window.onlaod=fu ...

  8. 正确使用 Volatile 变量——Brian Goetz

    本文转自:http://www.ibm.com/developerworks/cn/java/j-jtp06197.html 由Java并发大师Brian Goetz所撰写的. Java 语言中的 v ...

  9. JS高级程序设计2nd部分知识要点4

    ECMAScript中所有函数的参数都是按值传递的. 5种基本数据类型: Undfined,Null,Boolean,Number,String. ECMAScript中的所有参数传递的都是值,不可能 ...

  10. 活动与服务onbind()