Favorite Donut

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 79    Accepted Submission(s): 17

Problem Description

Lulu has a sweet tooth. Her favorite food is ring donut. Everyday she buys a ring donut from the same bakery. A ring donut is consists of n parts. Every part has its own sugariness that can be expressed by a letter from a to z (from low to high), and a ring donut can be expressed by a string whose i-th character represents the sugariness of the i−th part in clockwise order. Note that z is the sweetest, and two parts are equally sweet if they have the same sugariness.

Once Lulu eats a part of the donut, she must continue to eat its uneaten adjacent part until all parts are eaten. Therefore, she has to eat either clockwise or counter-clockwise after her first bite, and there are 2n ways to eat the ring donut of n parts. For example, Lulu has 6 ways to eat a ring donut abc: abc,bca,cab,acb,bac,cba. Lulu likes eating the sweetest part first, so she actually prefer the way of the greatest lexicographic order. If there are two or more lexicographic maxima, then she will prefer the way whose starting part has the minimum index in clockwise order. If two ways start at the same part, then she will prefer eating the donut in clockwise order. Please compute the way to eat the donut she likes most.

Input
First line contain one integer T,T≤20, which means the number of test case.

For each test case, the first line contains one integer n,n≤20000, which represents how many parts the ring donut has. The next line contains a string consisted of n lowercase alphabets representing the ring donut.

Output
You should print one line for each test case, consisted of two integers, which represents the starting point (from 1 to n) and the direction (0 for clockwise and 1 for counterclockwise).

Sample Input
2
4
abab
4
aaab
 
Sample Output
2 0
4 0
 
Source

解题:后缀自动机

 #include <bits/stdc++.h>
using namespace std;
const int maxn = ;
struct SAM {
struct node {
int son[],f,len;
void init() {
f = -;
len = ;
memset(son,-,sizeof son);
}
} sn[maxn<<];
int tot,last;
void init() {
tot = last = ;
sn[tot++].init();
}
int newnode() {
sn[tot].init();
return tot++;
}
void extend(int c) {
int np = newnode(),p = last;
sn[np].len = sn[last].len + ;
while(p != - && sn[p].son[c] == -) {
sn[p].son[c] = np;
p = sn[p].f;
}
if(p == -) sn[np].f = ;
else {
int q = sn[p].son[c];
if(sn[p].len + == sn[q].len) sn[np].f = q;
else {
int nq = newnode();
sn[nq] = sn[q];
sn[nq].len = sn[p].len + ;
sn[np].f = sn[q].f = nq;
while(p != - && sn[p].son[c] == q) {
sn[p].son[c] = nq;
p = sn[p].f;
}
}
}
last = np;
}
} sam;
char str[maxn];
int fail[maxn];
void getFail(const char *word) {
fail[] = -;
fail[] = ;
for(int i = ,j = -; word[i]; ++i) {
while(j != - && word[i] != word[j]) j = fail[j];
fail[i+] = ++j;
}
}
char text[maxn];
int mymin(int x,int y) {
return x > y?y:x;
}
int mymax(int x,int y) {
return x > y?x:y;
}
int FindIndex(const char *word,int (*f)(int,int),int idx,int len) {
for(int i = , j = ; text[i] ; ++i) {
while(j > - && word[j] != text[i]) j = fail[j];
if(!word[++j]) {
if(i - len + <= len)idx = f(idx,i - len + );
j = fail[j];
}
}
return idx;
}
int main() {
int kase,len;
scanf("%d",&kase);
while(kase--) {
sam.init();
scanf("%d",&len);
scanf("%s",str);
int p = ;
for(int i = ; i < (len<<); ++i)
sam.extend(str[i%len]-'a');
string ss = "",sb = "";
for(int i = ; i < len; ++i) {
for(int j = ; j >= ; --j) {
if(sam.sn[p].son[j] != -) {
p = sam.sn[p].son[j];
ss += char('a' + j);
break;
}
}
}
int clockwise = sam.sn[p].len - len + ;
getFail(ss.c_str());
strcpy(text,str);
strcpy(text + len,str);
clockwise = FindIndex(ss.c_str(),mymin,len*,len);
reverse(str,str + len);
sam.init();
for(int i = p = ; i < (len<<); ++i) {
sam.extend(str[i%len]-'a');
}
for(int i = ; i < len; ++i) {
for(int j = ; j >= ; --j) {
if(sam.sn[p].son[j] != -) {
p = sam.sn[p].son[j];
sb += char('a' + j);
break;
}
}
}
getFail(sb.c_str());
strcpy(text,str);
strcpy(text + len,str);
int anticlockwise = len - FindIndex(sb.c_str(),mymax,,len) + ;
if(ss == sb) {
if(clockwise == anticlockwise) {
printf("%d %d\n",clockwise,);
} else if(clockwise < anticlockwise) {
printf("%d %d\n",clockwise,);
} else printf("%d %d\n",anticlockwise,);
} else if(ss < sb) printf("%d %d\n",anticlockwise,);
else printf("%d %d\n",clockwise,);
}
return ;
}

HDU 5442 Favorite Donut的更多相关文章

  1. Hdu 5442 Favorite Donut (2015 ACM/ICPC Asia Regional Changchun Online 最大最小表示法 + KMP)

    题目链接: Hdu 5442 Favorite Donut 题目描述: 给出一个文本串,找出顺时针或者逆时针循环旋转后,字典序最大的那个字符串,字典序最大的字符串如果有多个,就输出下标最小的那个,如果 ...

  2. hdu 5442 Favorite Donut 后缀数组

    Favorite Donut Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid ...

  3. HDU 5442 Favorite Donut(暴力 or 后缀数组 or 最大表示法)

    http://acm.hdu.edu.cn/showproblem.php?pid=5442 题意:给出一串字符串,它是循环的,现在要选定一个起点,使得该字符串字典序最大(顺时针和逆时针均可),如果有 ...

  4. HDU 5442——Favorite Donut——————【最大表示法+kmp | 后缀数组】

    Favorite Donut Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) ...

  5. hdu 5442 Favorite Donut 最大表示法+kmp

    题目链接 给你一个字符串, 然后把他想象成一个环. 从某一个地方断开,然后逆时针或顺时针, 都可以形成一个字符串, 求字典序最大的那种. 输出断开位置以及是顺时针还是逆时针. 如果两个一样, 输出位置 ...

  6. Favorite Donut(HDU 5442)最小表示法+二分

    题目给出一个字符串,由a~z表示甜度,随字典序增大,字符串首尾相连形成一个圈,要求从一个位置开始字典序最大的字符串,输出位置以及是顺时针还是逆时针表示.顺时针用0表示,逆时针用1表示. 此题只需要查找 ...

  7. 【HDU - 5442】Favorite Donut 【最大表示法+KMP/后缀数组】

    题意 给出一个长度为n的环状由小写字母组成的序列,请找出从何处断开,顺时针还是逆时针,使得字典序最大.如果两个字符串的字典序一样大,那么它会选择下下标最小的那个.如果某个点顺时针逆时针产生的字典序大小 ...

  8. HDU 5442 后缀自动机(从环字符串选定一个位置 , 时针或顺时针走一遍,希望得到字典序最大)

    http://acm.hdu.edu.cn/showproblem.php?pid=5442 题目大意: 给定一个字符串,可理解成环,然后选定一位置,逆时针或顺时针走一遍,希望得到字典序最大,如果同样 ...

  9. hdu 5442 (ACM-ICPC2015长春网络赛F题)

    题意:给出一个字符串,长度是2*10^4.将它首尾相接形成环,并在环上找一个起始点顺时针或逆时针走一圈,求字典序最大的走法,如果有多个答案则找起始点最小的,若起始点也相同则选择顺时针. 分析:后缀数组 ...

随机推荐

  1. 回顾2017Java 小结

    一.Java语言最流行 最近,调查结果已公布:Java 被评为最流行的语言,JavaScript 是最常用的语言,而 Go 被认为是最有前途的语言,Python 是最多人想去尝试的语言. https: ...

  2. LuoguP4462 [CQOI2018]异或序列

    https://zybuluo.com/ysner/note/1124952 题面 给你一个大小为\(n\)的序列,然后给你一个数字\(k\),再给出\(m\)组询问,询问给出一个区间,问这个区间里面 ...

  3. 使用display:flex;实现垂直水平居中

    body,div{margin:0px;padding:0px;} .flex-container{display:flex;height:300px;background-color:#ddd;ju ...

  4. Maya Calendar

    http://poj.org/problem?id=1008 按第一种记录方法算出总天数,然后按第二种记录方式输出. #include<stdio.h> #include<strin ...

  5. markdownpad2下载安装教程

    1.下载安装 http://markdownpad.com/download/markdownpad2-setup.exe 直接下载,安装过程中提醒要安装微软的一个什么环境,不用理会直接跳过,实测没有 ...

  6. 马拉车算法(Manacher's Algorithm)

    这是悦乐书的第343次更新,第367篇原创 Manacher's Algorithm,中文名叫马拉车算法,是一位名叫Manacher的人在1975年提出的一种算法,解决的问题是求最长回文子串,神奇之处 ...

  7. SpringCloud服务组合

    SpringCloud生态强调微服务,微服务也就意味着将各个功能独立的业务抽象出来,做成一个单独的服务供外部调用.但每个人对服务究竟要有多“微”的理解差异很大,导致微服务的粒度很难掌控,划分规则也不统 ...

  8. Sublime Text Version 3.0,Build3143注册码

    1.打开sublime text软件2.Help->Enter License3.复制以下BEGIN LICENSE和END LICENSE之间的部分,粘贴进去.(注意:不要复制BEGIN LI ...

  9. 【Leetcode146】LRU Cache

    问题描述: 设计一个LRU Cache . LRU cache 有两个操作函数. 1.get(key). 返回cache 中的key对应的 val 值: 2.set(key, value). 用伪代码 ...

  10. debug时红点消失

    问题描述:debug时红色断点和黄色小箭头不见,而用行代码高亮的形式时. 解决办法:可以用设置 工具 => 选项 => 文本编辑器 => 指示器边距 勾上选项