[SPOJ-BEADS]Glass Beads

来源:
  CE1998

题目大意:
  求字符串最小表示。

思路:
  字符串复制一遍接在后面,构建SAM,然后每次跑小的转移。
  跑n次以后就跑到了最小表示的末尾,用该状态的len值减去n就是最小表示的起始位置。

 1 #include<string>
 2 #include<iostream>
 3 int n;
 4 std::string s;
 5 class SuffixAutomaton {
 6     private:
 7         static const int SIGMA_SIZE=26;
 8         struct State {
 9             State *link,*trans[SIGMA_SIZE];
10             unsigned len,min_trans;
11             State(const int l) {
12                 link=nullptr;
13                 std::fill(&trans[0],&trans[SIGMA_SIZE],nullptr);
14                 min_trans=SIGMA_SIZE;
15                 len=l;
16             }
17         };
18         State *root,*last;
19         unsigned idx(const int ch) {
20             return ch-‘a‘;
21         }
22         void extend(const char ch) {
23             const unsigned w=idx(ch);
24             State *p=last,*new_p=new State(last->len+1);
25             while(p!=nullptr&&p->trans[w]==nullptr) {
26                 p->trans[w]=new_p;
27                 p->min_trans=std::min(p->min_trans,w);
28                 p=p->link;
29             }
30             if(p==nullptr) {
31                 new_p->link=root;
32             } else {
33                 State *q=p->trans[w];
34                 if(q->len==p->len+1) {
35                     new_p->link=q;
36                 } else {
37                     State *new_q=new State(p->len+1);
38                     std::copy(&q->trans[0],&q->trans[SIGMA_SIZE],new_q->trans);
39                     new_q->min_trans=q->min_trans;
40                     new_q->link=q->link;
41                     q->link=new_p->link=new_q;
42                     while(p!=nullptr&&p->trans[w]==q) {
43                         p->trans[w]=new_q;
44                         p=p->link;
45                     }
46                 }
47             }
48             last=new_p;
49         }
50     public:
51         void build() {
52             root=last=new State(0);
53             for(std::string::iterator i=s.begin();i<s.end();i++) {
54                 extend(*i);
55             }
56             for(std::string::iterator i=s.begin();i<s.end();i++) {
57                 extend(*i);
58             }
59         }
60         unsigned query() {
61             State *p=root;
62             for(unsigned i=0;i<s.length();i++) {
63                 p=p->trans[p->min_trans];
64             }
65             return p->len-s.length()+1;
66         }
67 };
68 SuffixAutomaton sam;
69 int main() {
70     std::ios_base::sync_with_stdio(false);
71     std::cin.tie(NULL);
72     std::cin>>n;
73     for(int i=0;i<n;i++) {
74         std::cin>>s;
75         sam.build();
76         std::cout<<sam.query()<<std::endl;
77     }
78     return 0;
79 }
时间: 2024-07-31 15:37:54

[SPOJ-BEADS]Glass Beads的相关文章

UVALive 5545 Glass Beads

Glass Beads Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 554564-bit integer IO format: %lld      Java class name: Main Once upon a time there was a famous actress. As you may expect, she played mostly

[最小表示] poj 1509 Glass Beads

题目链接: http://poj.org/problem?id=1509 Glass Beads Time Limit: 3000MS   Memory Limit: 10000K Total Submissions: 2311   Accepted: 1343 Description Once upon a time there was a famous actress. As you may expect, she played mostly Antique Comedies most of

Glass Beads

Once upon a time there was a famous actress. As you may expect, she played mostly Antique Comedies most of all. All the people loved her. But she was not interested in the crowds. Her big hobby were beads of any kind. Many bead makers were working fo

PKU 1509 Glass Beads (最小表示法)

题意:有一个环形字符串,让你找一个位置切一刀使得字符串字母序最小,输出这个位置. 思路:可以看成两个字符串比较,一个是从下标0开始(0~n-1),一个从下标1开始(1~n-1,0). 然后两个指针i=0,j=1.从s[i]和s[j]开始比较第k个字符是否相同,当k==len时,返回i,j中的最小值.当s[i+k]和s[j+k]不相同时,若s[i+k]>s[j+k]则可见从s[i+1]到s[i+k]都不会是最小字典序的起始位置,所以i=i+k+1.当s[i+k]<s[j+k]时同理.若移动后i=

zoj 2006 Glass Beads

Glass Beadshttp://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1006 Time Limit: 2 Seconds      Memory Limit: 65536 KB Once upon a time there was a famous actress. As you may expect, she played mostly Antique Comedies most of all. All the peopl

SPOJ-BEADS UVA719 UVALive5545 POJ1509 ZOJ2006 Glass Beads【字符串环的最小】

Glass Beads Time Limit: 3000MS Memory Limit: 10000K Total Submissions: 5254 Accepted: 2943 Description Once upon a time there was a famous actress. As you may expect, she played mostly Antique Comedies most of all. All the people loved her. But she w

poj 1509 Glass Beads

题意:给你一个长度为n的字符串环,以位置i开始的顺时针长度为n的环构成的字符串有n个,问其中最小字典序的开始位置,有多种解时,输出起始位置最小的. 分析: 首先可以直接拼接两个长度为n的字符串,设原串为S[0],S[1]...S[n-1]则拼接后就是S'=S[0],S[1],...S[n-1],S[0],S[1],...S[n-1]. 那么问题中的n个长度为n的字符串中的任意一个,一定存在S'的某个后缀字符串的前缀与其相等. 我们现在要找最小字典序,则可以直接先求S'的后缀数组SA,然后: 1.

uva 719 Glass Beads(后缀自动机)

[题目链接] https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=524&page=show_problem&problem=660 [题目大意] 给出一个字符串,求出与其循环同构的字符串中,字典序最小的一个. [题解] 以原字符串的两倍建立自动机,按字典序在parent树上搜索, 得到的第一个长度为n的字符串就是答案. [代码] #include <cstdio

杂项(最小表示法):HZOI 2015 Glass Beads

[题目描述] 给定长度为n(n<=300000)的循环同构的字符串,定义最小表示为该字符串的字典序最小的同构表示,请输出这个表示. [输入格式] 第一行是串的长度,第二行是字符串. [输出格式] 串的最小表示. [样例输入] 10 helloworld [样例输出] dhelloworl [题目来源] HZOI2015 改编自poj1509 算法很显然,需要注意的是:s[len+1]要赋值成一个较大值. 1 #include <iostream> 2 #include <cstri

POJ 1509 Glass Beads 后缀自动机

题目大意:给出一个环形的字符串,问从哪里开始是的这个字符串的字典序最小. 思路:最小表示法和后缀自动机的裸题,不过我是为了学后缀自动机才写的这个题,就没有去学最小表示法. 做法很简单,先建立一个后缀自动机,然后从根开始沿tranc指针从a->z走len次到达的点就是字典序最小的字符串的结尾点,求起始点只要减一下长度再+1即可. 对于后缀自动机的理解:http://wyfcyx.is-programmer.com/posts/76107.html CODE: #include <cstdio&g