Wireless Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5640    Accepted Submission(s): 1785 Problem Description Liyuan lives in a old apartment. One day, he suddenly found that there…

题目链接 BZOJ1559 题解 考虑到这是一个包含子串的问题,而且子串非常少,我们考虑\(AC\)自动机上的状压\(dp\) 设\(f[i][j][s]\)表示长度为\(i\)的串,匹配到了\(AC\)自动机\(j\)号节点,且已匹配集合为\(s\)的方案数 直接在\(AC\)自动机上转移即可 但是为了防止使用\(last\)指针之类的,计算匹配的串,我们先将原串的集合去重和去包含关系 方案怎么办? 考虑到\(ans \le 42\),一定是刚好若干个原串以最长前后缀相同的方式相接 因为如果不…

题意:目标串n( <= 10)个,病毒串m( < 1000)个,问包含所有目标串无病毒串的最小长度 思路:貌似是个简单的状压DP + AC自动机,但是发现dp[1 << n][5e4]根本开不出那么多空间,似乎GG.但是我们仔细想一下就能发现,既然要包含所有目标串的最小长度,那必然这个串就是只有目标串叠加组成的,只是在叠加的过程中我们不能混入病毒串.所以其实Trie树上有用的点最多就10个,我们只要处理出所有目标串之间"最小有效转化"就行了,那么空间为dp[1…

Time Limit: 10 Seconds      Memory Limit: 65536 KB Dr. X is a biologist, who likes rabbits very much and can do everything for them. 2012 is coming, and Dr. X wants to take some rabbits to Noah's Ark, or there are no rabbits any more. A rabbit's gene…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5400    Accepted Submission(s): 1704 Problem Description Liyuan lives in a old apartment. One day, he suddenly found that there was a wireless ne…

题目链接 Problem Description Dr. X is a biologist, who likes rabbits very much and can do everything for them. 2012 is coming, and Dr. X wants to take some rabbits to Noah's Ark, or there are no rabbits any more. A rabbit's genes can be expressed as a st…

题目,求包含所有的给定的n个DNA片段的序列的最短长度. AC自动机上的DP题. dp[S][u]表示已经包含的DNA片段集合为S,且当前后缀状态是自动机第u个结点的最短长度 dp[0][0]=0 我为人人+队列轻松转移.. #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; #define INF (1<<30)…

题目链接 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has n 01 strings si, and he wants to know the number of 01 antisymmetric stri…

给你m个字符串,让你构造一个字符串,包含所有的m个子串,问有多少种构造方法.如果答案不超过42,则按字典序输出所有可行解. 由于m很小,所以可以考虑状压. 首先对全部m个子串构造出AC自动机,每个节点有一个附加属性val[u]代表结点u包含的子串集合. 设dp[l][S][u]为长度为l,包含子串集合为S,当前在结点u时,接下来能构造出的合法字符串总数,则dp[l][S][u]=∑dp[l+1][S|val[v]][v],v=go[u][i],0<=i<26; 两遍dfs,一遍dp,一遍输出答…

传送门 题目大意: 给你一些密码片段字符串,让你求长度为n,且至少包含k个不同密码片段串的字符串的数量. 题解: 因为密码串不多,可以考虑状态压缩 设dp[i][j][sta]表示长为i的字符串匹配到j节点且状态为sta的数量. 其中sta存储的是包含的密码串情况,在构建fail指针时,当前节点要并上fail指针所指的节点. 跑ac自动机,儿子节点从父亲节点转移. 最后取dp[len][...][sta]的和,其中sta满足二进制中1的数量>=k, 这一点可以像树状数组的lowbit那样快速求出…