题目大意: 用集合A中的串构造出一个串,使之让很多其它的setB中的串成为他的子串. 思路分析: 和 Codeforces 86C 几乎相同. 只是这里是要用A中的构造. 先用A 和 B的串构造一个自己主动机.然后对于A集合的尾结点给出一个最大后缀匹配,对于B集合的尾结点给一个权值. dp[i][j][k] 表示已经构造出来了一个长度为i的串,如今走到了自己主动机的j结点.i长度后面有k个字符是没有匹配到的. 继续在自己主动机上走进行状态转移. if(isword >= k +1 )dp [i+…

题目链接:Searching the String 解析:给一个长串.给n个不同种类的短串.问分别在能重叠下或者不能重叠下短串在长串中出现的次数. 能重叠的已经是最简单的AC自己主动机模板题了. 不能重叠的记录一下每一个匹配的串的起始位置保证不重叠就可以. AC代码: #include <bits/stdc++.h> using namespace std; struct Trie{ int next[600010][26], fail[600010], deep[600010]; int r…

Detect the Virus Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita found that his computer is extremely slow. After several hours' work, he finally found that it was a virus that made his poor computer slow and the virus was activated…

DNA Sequence 题意:DNA的序列由ACTG四个字母组成,如今给定m个不可行的序列.问随机构成的长度为n的序列中.有多少种序列是可行的(仅仅要包括一个不可行序列便不可行).个数非常大.对100000取模. 思路:推荐一个博客,讲的很清楚. 这样的题目.n非常大,首先想到的就是用矩阵来优化.那么怎样构造转移方程呢:首先建立一棵Trie,然后依照AC自己主动机的方式构造fail指针,然后会发现.当一个状态分别加入ACTG之后,会得到还有一个状态. (详细解释见代码) 代码: /* ID:…

Walk Through Squares Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 944 Accepted Submission(s): 277 Problem Description On the beaming day of 60th anniversary of NJUST, as a military college whic…

DNA repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 5877   Accepted: 2760 Description Biologists finally invent techniques of repairing DNA that contains segments causing kinds of inherited diseases. For the sake of simplicity, a…

Rescue the Rabbit Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1412 Accepted Submission(s): 403 Problem Description Dr. X is a biologist, who likes rabbits very much and can do everything for…

Description   On the beaming day of 60th anniversary of NJUST, as a military college which was Second Artillery Academy of Harbin Military Engineering Institute before, queue phalanx is a special landscape.      Here is a M*N rectangle, and this one…

Description Liyuan lives in a old apartment. One day, he suddenly found that there was a wireless network in the building. Liyuan did not know the password of the network, but he got some important information from his neighbor. He knew the password…

标题效果: 举个很多种DNA弦,每个字符串值值至1.最后,一个长字符串.要安排你最后一次另一个字符串,使其没事子值和最大. IDEAS: 首先easy我们的想法是想搜索的!管她3721..直接一个字符一个字符的码,然后在AC自己主动机上推断最后的权值.TLE哟. 然后发现搜过不去.那就dp咯.再easy想到的就是dp[i][a][b][c][d] 表示此时遍历AC自己主动机的节点在i,然后构成了a个A,b个G,c个C.d个T的权值. 再一看内存,500*40*40*40*40...然后...就没…