You are the King of Byteland. Your agents have just intercepted a batch of encrypted enemy messages concerning the date of the planned attack on your island. You immedietaly send for the Bytelandian Cryptographer, but he is currently busy eating popcorn and claims that he may only decrypt the most important part of the text (since the rest would be a waste of his time). You decide to select the fragment of the text which the enemy has strongly emphasised, evidently regarding it as the most important. So, you are looking for a fragment of text which appears in all the messages disjointly at least twice. Since you are not overfond of the cryptographer, try to make this fragment as long as possible.

Input

The first line of input contains a single positive integer t<=10, the number of test cases. t test cases follow. Each test case begins with integer n (n<=10), the number of messages. The next n lines contain the messages, consisting only of between 2 and 10000 characters 'a'-'z', possibly with some additional trailing white space which should be ignored.

Output

For each test case output the length of longest string which appears disjointly at least twice in all of the messages.

题目大意:给n个字符串,求在每个字符串中出现至少两次且不重叠的最长子串的长度。

思路:每个字符串用不相同的字符连起来。求后缀数组和height[]数组。

二分长度L,检查长度。

检查的时候从前往后扫一遍,看有没有连起来的一部分,公共前缀大于等于L且在每个字符串中都出现了且不重叠。

复杂度为O(nlogn)

代码(0.34s):

 #include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std; const int MAXN = ; char s[MAXN];
int id[MAXN];
int sa[MAXN], rank[MAXN], height[MAXN], c[MAXN], tmp[MAXN];
int n, m, T; void makesa(int m) {
memset(c, , m * sizeof(int));
for(int i = ; i < n; ++i) ++c[rank[i] = s[i]];
for(int i = ; i < m; ++i) c[i] += c[i - ];
for(int i = ; i < n; ++i) sa[--c[rank[i]]] = i;
for(int k = ; k < n; k <<= ) {
for(int i = ; i < n; ++i) {
int j = sa[i] - k;
if(j < ) j += n;
tmp[c[rank[j]]++] = j;
}
int j = c[] = sa[tmp[]] = ;
for(int i = ; i < n; ++i) {
if(rank[tmp[i]] != rank[tmp[i - ]] || rank[tmp[i] + k] != rank[tmp[i - ] + k])
c[++j] = i;
sa[tmp[i]] = j;
}
memcpy(rank, sa, n * sizeof(int));
memcpy(sa, tmp, n * sizeof(int));
}
} void calheight() {
for(int i = , k = ; i < n; height[rank[i++]] = k) {
k -= (k > );
int j = sa[rank[i] - ];
while(s[i + k] == s[j + k]) ++k;
}
} int mx[MAXN], mn[MAXN];
int stk[MAXN]; void update_max(int &a, int b) {
if(a == - || a < b) a = b;
} void update_min(int &a, int b) {
if(a == - || a > b) a = b;
} bool check(int L) {
int sum = , top = ;
memset(mx, -, m * sizeof(int));
memset(mn, -, m * sizeof(int));
memset(c, , m * sizeof(int));
for(int i = ; i < n; ++i) {
if(height[i] >= L) {
update_max(mx[id[sa[i]]], sa[i]);
update_min(mn[id[sa[i]]], sa[i]);
stk[++top] = id[sa[i]];
if(mx[id[sa[i]]] - mn[id[sa[i]]] >= L) {
if(!c[id[sa[i]]]) ++sum;
c[id[sa[i]]] = true;
if(sum >= m) return true;
}
} else {
sum = ;
while(top) {
int t = stk[top--];
mx[t] = mn[t] = -;
c[t] = false;
}
update_max(mx[id[sa[i]]], sa[i]);
update_min(mn[id[sa[i]]], sa[i]);
stk[++top] = id[sa[i]];
}
}
return false;
} int solve() {
int l = , r = ;
while(l < r) {
int mid = (l + r) >> ;
if(check(mid)) l = mid + ;
else r = mid;
}
return l - ;
} int main() {
scanf("%d", &T);
while(T--) {
scanf("%d", &m);
n = ;
for(int i = ; i < m; ++i) {
scanf("%s", s + n);
while(s[n]) id[n++] = i;
s[n++] = i + ;
}
s[n - ] = ;
makesa();
calheight();
printf("%d\n", solve());
}
}

SPOJ 220 Relevant Phrases of Annihilation(后缀数组)的更多相关文章

  1. SPOJ - PHRASES Relevant Phrases of Annihilation —— 后缀数组 出现于所有字符串中两次且不重叠的最长公共子串

    题目链接:https://vjudge.net/problem/SPOJ-PHRASES PHRASES - Relevant Phrases of Annihilation no tags  You ...

  2. SPOJ 220 Relevant Phrases of Annihilation(后缀数组+二分答案)

    [题目链接] http://www.spoj.pl/problems/PHRASES/ [题目大意] 求在每个字符串中出现至少两次的最长的子串 [题解] 注意到这么几个关键点:最长,至少两次,每个字符 ...

  3. SPOJ220 Relevant Phrases of Annihilation(后缀数组)

    引用罗穗骞论文中的话: 先将n 个字符串连起来,中间用不相同的且没有出现在字符串中的字符隔开,求后缀数组.然后二分答案,再将后缀分组.判断的时候,要看是否有一组后缀在每个原来的字符串中至少出现两次,并 ...

  4. 【SPOJ 220】Relevant Phrases of Annihilation

    http://www.spoj.com/problems/PHRASES/ 求出后缀数组然后二分. 因为有多组数据,所以倍增求后缀数组时要特判是否越界. 二分答案时的判断要注意优化! 时间复杂度\(O ...

  5. POJ - 3294~Relevant Phrases of Annihilation SPOJ - PHRASES~Substrings POJ - 1226~POJ - 3450 ~ POJ - 3080 (后缀数组求解多个串的公共字串问题)

    多个字符串的相关问题 这类问题的一个常用做法是,先将所有的字符串连接起来, 然后求后缀数组 和 height 数组,再利用 height 数组进行求解. 这中间可能需要二分答案. POJ - 3294 ...

  6. SPOJ - PHRASES K - Relevant Phrases of Annihilation

    K - Relevant Phrases of Annihilation 题目大意:给你 n 个串,问你最长的在每个字符串中出现两次且不重叠的子串的长度. 思路:二分长度,然后将height分块,看是 ...

  7. SPOJ - PHRASES Relevant Phrases of Annihilation (后缀数组)

    You are the King of Byteland. Your agents have just intercepted a batch of encrypted enemy messages ...

  8. SPOJ PHRASES Relevant Phrases of Annihilation(后缀数组 + 二分)题解

    题意: 给\(n\)个串,要你求出一个最长子串\(A\),\(A\)在每个字串至少都出现\(2\)次且不覆盖,问\(A\)最长长度是多少 思路: 后缀数组处理完之后,二分这个长度,可以\(O(n)\) ...

  9. 【SPOJ 220】 PHRASES - Relevant Phrases of Annihilation

    [链接]h在这里写链接 [题意]     给你n(n<=10)个字符串.     每个字符串长度最大为1e4;     问你能不能找到一个子串.     使得这个子串,在每个字符串里面都不想交出 ...

随机推荐

  1. BLE Device Monitor的使用

    1 综述 BLE Device Monitor是一个用来显示任意蓝牙低功耗设备服务(services).特征(characteristics).属性(attributes)的windows程序.除了测 ...

  2. WGZX:javaScript 学习心得--2

    转贴javascript心得(二) 标签: javascriptajaxweb开发htmlfirefox框架 2008-09-11 10:56 636人阅读 评论(0) 收藏 举报  分类: UI(2 ...

  3. Java高级之虚拟机垃圾回收机制

    博客出自:http://blog.csdn.net/liuxian13183,转载注明出处! All Rights Reserved ! 区别于C语言手动回收,Java自动执行垃圾回收,但为了执行高效 ...

  4. PHP---关联模型

    MANY_TO_MANY

  5. python StringIO

    模块是用类编写的,只有一个StringIO类,所以它的可用方法都在类中. 此类中的大部分函数都与对文件的操作方法类似. 例: 复制代码 代码如下: #coding=gbk   import Strin ...

  6. Rails进阶参考

    https://gist.github.com/xdite/4044f3a037de029bc35c From idea to products: - Ideation, wireframes, mo ...

  7. ionic build android 报错分析

  8. Java文件操作①——XML文件的读取

    一.邂逅XML 文件种类是丰富多彩的,XML作为众多文件类型的一种,经常被用于数据存储和传输.所以XML在现今应用程序中是非常流行的.本文主要讲Java解析和生成XML.用于不同平台.不同设备间的数据 ...

  9. 使用sh-x调试shell脚本_转

    参考:http://blog.chinaunix.net/uid-20564848-id-73502.html 1. 通过sh -x 脚本名  #显示脚本执行过程2.脚本里set -x选项,轻松跟踪调 ...

  10. SQLdiag-配置文件-ProfilerCollector

    上一篇,我们讲述了配置文件中与性能计数器相关的PerfmonCollector元素:这一篇我们将讲述与跟踪数据相关的ProfilerCollector元素.在上一篇中使用SD_Detailed.XML ...