[POJ2774][codevs3160]Long Long Message
[POJ2774][codevs3160]Long Long Message
试题描述
The little cat lives in an unrich family, so he frequently comes to the mobile service center, to check how much money he has spent on SMS. Yesterday, the computer of service center was broken, and printed two very long messages. The brilliant little cat soon found out:
1. All characters in messages are lowercase Latin letters, without punctuations and spaces.
2. All SMS has been appended to each other – (i+1)-th SMS comes directly after the i-th one – that is why those two messages are quite long.
3. His own SMS has been appended together, but possibly a great many redundancy characters appear leftwards and rightwards due to the broken computer.
E.g: if his SMS is “motheriloveyou”, either long message printed by that machine, would possibly be one of “hahamotheriloveyou”, “motheriloveyoureally”, “motheriloveyouornot”, “bbbmotheriloveyouaaa”, etc.
4. For these broken issues, the little cat has printed his original text twice (so there appears two very long messages). Even though the original text remains the same in two printed messages, the redundancy characters on both sides would be possibly different.
You are given those two very long messages, and you have to output the length of the longest possible original text written by the little cat.
Background:
The SMS in Byterland mobile service are charging in dollars-per-byte. That is why the little cat is worrying about how long could the longest original text be.
Why ask you to write a program? There are four resions:
1. The little cat is so busy these days with physics lessons;
2. The little cat wants to keep what he said to his mother seceret;
3. POJ is such a great Online Judge;
4. The little cat wants to earn some money from POJ, and try to persuade his mother to see the doctor :(
输入
Two strings with lowercase letters on two of the input lines individually. Number of characters in each one will never exceed 100000.
输出
A single line with a single integer number – what is the maximum length of the original text written by the little cat.
输入示例
- yeshowmuchiloveyoumydearmotherreallyicannotbelieveit
- yeaphowmuchiloveyoumydearmother
输出示例
数据规模及约定
见“输入”
题解
两个串拼在一起(中间别忘了加分隔符)跑一边后缀排序求出 height 数组,然后二分答案,对于 height[i] ≥ x 的分为一组,然后 check 时看一组内是否两个串的后缀都出现就行了。
- #include <iostream>
- #include <cstdio>
- #include <cstdlib>
- #include <cstring>
- #include <cctype>
- #include <algorithm>
- using namespace std;
- #define maxn 200010
- char S[maxn], S2[maxn];
- int n, rank[maxn], height[maxn], sa[maxn], Ws[maxn];
- bool cmp(int* a, int p1, int p2, int l) {
- if(p1 + l > n && p2 + l > n) return a[p1] == a[p2];
- if(p1 + l > n || p2 + l > n) return 0;
- return a[p1] == a[p2] && a[p1+l] == a[p2+l];
- }
- void ssort() {
- int *x = rank, *y = height;
- int m = 0;
- for(int i = 1; i <= n; i++) Ws[x[i] = S[i]]++, m = max(m, x[i]);
- for(int i = 1; i <= m; i++) Ws[i] += Ws[i-1];
- for(int i = n; i; i--) sa[Ws[x[i]]--] = i;
- for(int j = 1, pos = 0; pos < n; j <<= 1, m = pos) {
- pos = 0;
- for(int i = n - j + 1; i <= n; i++) y[++pos] = i;
- for(int i = 1; i <= n; i++) if(sa[i] > j) y[++pos] = sa[i] - j;
- for(int i = 1; i <= m; i++) Ws[i] = 0;
- for(int i = 1; i <= n; i++) Ws[x[i]]++;
- for(int i = 1; i <= m; i++) Ws[i] += Ws[i-1];
- for(int i = n; i; i--) sa[Ws[x[y[i]]]--] = y[i];
- swap(x, y); pos = 1; x[sa[1]] = 1;
- for(int i = 2; i <= n; i++) x[sa[i]] = cmp(y, sa[i], sa[i-1], j) ? pos : ++pos;
- }
- return ;
- }
- void calch() {
- for(int i = 1; i <= n; i++) rank[sa[i]] = i;
- for(int i = 1, j, k = 0; i <= n; height[rank[i++]] = k)
- for(k ? k-- : 0, j = sa[rank[i]-1]; S[j+k] == S[i+k]; k++);
- return ;
- }
- bool check(int x, int p) {
- bool fl = 0, fr = 0;
- if(sa[1] <= p) fl = 1; else fr = 1;
- for(int i = 2; i <= n; i++) {
- if(height[i] < x) {
- fl = fr = 0;
- if(sa[i] <= p) fl = 1; else fr = 1;
- }
- else {
- if(sa[i] <= p) fl = 1; else fr = 1;
- if(fl & fr) return 1;
- }
- }
- return 0;
- }
- int main() {
- scanf("%s%s", S + 1, S2 + 1);
- n = strlen(S + 1); int m = strlen(S2 + 1);
- for(int i = 1; i <= m; i++) S[n+i+1] = S2[i];
- S[n+1] = '#';
- swap(m, n);
- n += m + 1;
- ssort();
- calch();
- int l = 0, r = n + 1;
- while(r - l > 1) {
- int mid = l + r >> 1;
- if(check(mid, m)) l = mid; else r = mid;
- }
- printf("%d\n", l);
- return 0;
- }
建出第一个串的 SAM 后用第二个串在上面匹配。
- #include <iostream>
- #include <cstdio>
- #include <cstdlib>
- #include <cstring>
- #include <cctype>
- #include <algorithm>
- using namespace std;
- #define maxn 200010
- #define maxa 26
- char S[maxn];
- int len;
- int rt, ToT, last, to[maxn][maxa], par[maxn], Max[maxn];
- void extend(int x) {
- int p = last, np = ++ToT; Max[np] = Max[p] + 1; last = np;
- while(p && !to[p][x]) to[p][x] = np, p = par[p];
- if(!p){ par[np] = rt; return ; }
- int q = to[p][x];
- if(Max[q] == Max[p] + 1){ par[np] = q; return ; }
- int nq = ++ToT; Max[nq] = Max[p] + 1;
- memcpy(to[nq], to[q], sizeof(to[q]));
- par[nq] = par[q];
- par[q] = par[np] = nq;
- while(p && to[p][x] == q) to[p][x] = nq, p = par[p];
- return ;
- }
- int main() {
- rt = ToT = last = 1;
- scanf("%s", S);
- len = strlen(S);
- for(int i = 0; i < len; i++) extend(S[i] - 'a');
- scanf("%s", S);
- int p = rt, n = strlen(S), tmp = 0, ans = 0;
- for(int i = 0; i < n; i++) {
- int x = S[i] - 'a';
- if(to[p][x]) tmp++, p = to[p][x];
- else {
- while(p && !to[p][x]) tmp = Max[par[p]] + 1, p = par[p];
- if(!p) tmp = 0, p = 1;
- else p = to[p][x];
- }
- ans = max(ans, tmp);
- }
- printf("%d\n", ans);
- return 0;
- }
[POJ2774][codevs3160]Long Long Message的更多相关文章
- 【POJ2774】Long Long Message(后缀数组)
[POJ2774]Long Long Message(后缀数组) 题面 Vjudge Description Little cat在Byterland的首都读物理专业.这些天他收到了一条悲伤地信息:他 ...
- 【POJ2774】Long Long Message (后缀数组)
Long Long Message Description The little cat is majoring in physics in the capital of Byterland. A p ...
- 【poj2774】 Long Long Message
http://poj.org/problem?id=2774 (题目链接) 题意 给出两个只包含小写字母的字符串,求出最长连续公共子串. solution 第一次用后缀数组,感觉有点神...才发现原来 ...
- HiHocoder1415 : 后缀数组三·重复旋律3 & Poj2774:Long Long Message
题面 HiHocoder1415 Poj2774 Sol 都是求最长公共子串,\(hihocoder\)上讲的很清楚 把两个串拼在一起,中间用一个特殊字符隔开 那么答案就是排序后相邻两个不同串的后缀的 ...
- POJ2774:Long Long Message——题解
http://poj.org/problem?id=2774 给定两个字符串 A 和 B,求最长公共子串. 论文题,把两个串合并起来,比较两个串各自的后缀的height值取最大即可. #include ...
- 【POJ2774】Long Long Message(后缀数组求Height数组)
点此看题面 大致题意: 求两个字符串中最长公共子串的长度. 关于后缀数组 关于\(Height\)数组的概念以及如何用后缀数组求\(Height\)数组详见这篇博客:后缀数组入门(二)--Height ...
- 【POJ2774】Long Long Message (SA)
最长公共子串...两个字符串连在一起,中间放一个特殊字符隔开.求出height之后,枚举height,看两个后缀是不是分布于两段字符串..如果是,这个值就可以作为答案.取最大值即可. ; var c, ...
- 【poj2774】Long Long Message
用个分隔符将两个字符串连接起来,再用后缀数组求出height数组的值,找出一个height值最大并且i与i-1的sa值分别在两串字符中就好 #include<algorithm> #inc ...
- POJ2774:Long Long Message
问两个串的最长公共子串,n<=100000. SAM可以直接搞当然SA哈希都可以..类似于KMP的做法,如果沿parent边走要顺势修改匹配位置. #include<stdio.h> ...
随机推荐
- VS2010环境下.NET4.0中Tuple<T>的一个小BUG问题
启动一个桌面程序后,发现一个窗体cfdata=null, 执行时发生错误, 但是在初始化的时候,我明明是cfdata=new Cfdata();为什么会出现这个错误呢. 我开始跟踪,发现当执行cfda ...
- javascript中闭包与作用域的理解
很多js的框架与插件编写都用到了闭包,所以,阅读和掌握闭包很有必要.最近学习vue框架时,经常会猜想很多功能的native js实现,很多都应用到了闭包,闭包除了目前已知的一些特性,如:可以保持局部变 ...
- nginx负载均衡浅析
熟悉Nginx的小伙伴都知道,Nginx是一个非常好的负载均衡器.除了用的非常普遍的Http负载均衡,Nginx还可以实现Email,FastCGI的负载均衡,甚至可以支持基于Tcp/UDP协议的各种 ...
- PostgreSQL学习手册(五) 函数和操作符
PostgreSQL学习手册(五) 函数和操作符 一.逻辑操作符: 常用的逻辑操作符有:AND.OR和NOT.其语义与其它编程语言中的逻辑操作符完全相同. 二.比较操作符: 下面是Post ...
- 【译】x86程序员手册36-9.9异常汇总
9.9 Exception Summary 异常汇总 Table 9-6 summarizes the exceptions recognized by the 386. Table 9-6. Exc ...
- Adobe Dreamweaver CC 2014 代码颜色目录 dw
他的颜色代码配置文件,不在安装目录下,这让我好找啊~ C:\Users\Administrator\AppData\Roaming\Adobe\Dreamweaver CC 2014\zh_CN\Co ...
- ansible中yaml语法应用
4.yaml语法应用 ansible的playbook编写是yaml语言编写,掌握yaml语法是编写playbook的必要条件,格式要求和Python相似,具体教程参考如下 yaml语言教程 附上一个 ...
- echo追加字符串到文件末尾
1.覆盖 echo "string" > filename 2.追加 echo "string" >> filename
- CSS三栏布局的四种方法
总括: 不管是三栏布局还是两栏布局都是我们在平时项目里经常使用的,也许你不知道什么事三栏布局什么是两栏布局但实际已经在用,或许你知道三栏布局的一种或两种方法,但实际操作中也只会依赖那某一种方法,本文具 ...
- show()的方向
<!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content ...