Life Forms Description - You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have…

有一个细节不是特别懂,然后的话细节有点多,就是挺难发现的那一种,感谢大佬的博客 1470: 后缀数组4:Life Forms poj3294 时间限制: 1 Sec  内存限制: 128 MB提交: 112  解决: 35[提交] [状态] [讨论版] [命题人:admin] 题目描述 [问题描述] 求n个字符串(长度1000)的最长的一个子串,满足该子串在一半以上(不包括一半)的字符串中出现过,并输出该子串,如果有多个子串满足要求,则按字典序输出所有的子串:(全部都是小写字母) [输入格式]…

题目链接:https://vjudge.net/problem/POJ-3294 Life Forms Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 16905   Accepted: 4970 Description You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial t…

Life Forms Description You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have ge…

Life Forms Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 12483   Accepted: 3501 Description You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, e…

(累了,这题做了很久!) Life Forms Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 8683   Accepted: 2375 Description You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour,…

http://poj.org/problem?id=3294 (题目链接) 题意 给定 n 个字符串,求出现在不小于 k 个字符串中的最长子串. Solution 后缀数组论文题.. 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,将后缀分成若干组,判断每组的后缀是否出现在不小于 k 个的原串中.这个做法的时间复杂度为 O(nlogn). 要求字典序从小到大排序的多组解,同样是利用rank进行排序即可. 细节 strlen害死人→_→,好像默认的A…

题意: n<=100 len[i]<=1000 思路:这是一道论文题 ..]of longint; ch:..]of ansistring; n,n1,l,r,mid,last,i,j,m,len:longint; procedure swap(var x,y:longint); var t:longint; begin t:=x; x:=y; y:=t; end; function cmp(a,b,l:longint):boolean; begin exit((y[a]=y[b])and(y…

引用罗穗骞论文中的话: 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,用和例3 同样的方法将后缀分成若干组,判断每组的后缀是否出现在不小于k 个的原串中.这个做法的时间复杂度为O(nlogn). #include<cstdio> #include<iostream> #include<algorithm> #include<vector> #include<cstring> #include<…

[题目链接] http://poj.org/problem?id=3294 [题意] 多个字符串求出现超过R次的最长公共子串. [思路] 二分+划分height,判定一个组中是否包含不小于R个不同字符串的后缀. 需要注意的有: 1)      c[]尽量开大,字符范围为“偏移”之后的范围. 2)      用kase作为标记节省了每次开始新段需要清零的时间. 3)      因为height是sa[i]与sa[i-1]的关系,所以无论是在can的开始还是在新段开始都需要初始为一个串的情况. [代…