题目链接:https://vjudge.net/problem/HDU-2825 Wireless Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7733    Accepted Submission(s): 2509 Problem Description Liyuan lives in a old apartmen…

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…

题目大意:给一系列字符串,用小写字母构造出长度为n的至少包含k个字符串的字符串,求能构造出的个数. 题目分析:在AC自动机上走n步,至少经过k个单词节点,求有多少种走法. 代码如下: # include<iostream> # include<cstdio> # include<queue> # include<cstring> # include<algorithm> using namespace std; typedef long long…

题目链接 做题, #include <cstdio> #include <string> #include <cstring> using namespace std; #define MOD 20090717 ][]; ]; ]; ]; ][][]; ]; int t; void CL() { memset(trie,-,sizeof(trie)); memset(dp,,sizeof(dp)); memset(o,,sizeof(o)); t = ; } void…

Wireless Password Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u 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…

HDU2825 Wireless Password Problem 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 h…

题目问长度m不包含一些不文明单词的字符串有多少个. 依然是水水的AC自动机+DP..做完后发现居然和POJ2778是一道题,回过头来看都水水的... dp[i][j]表示长度i(在自动机转移i步)且后缀状态为自动机第j个结点的合法字符串数 dp[0][0]=1 转移转移... 注意要用高精度,因为答案最多5050. 还有就是要用unsigned char,题目的输入居然有拓展的ASCII码,编码128-255. #include<cstdio> #include<cstring>…

题目是给几个带有价值的单词.而一个字符串的价值是 各单词在它里面出现次数*单词价值 的和,问长度不超过n的最大价值的字符串是什么? 依然是入门的AC自动机+DP题..不一样的是这题要输出具体方案,加个字符数组记录每个状态最优情况的字符串即可. 另外题目字典序是先考虑长度再考虑每一位单词:特别要注意,有一个非常坑的地方看了Disscus才知道——单词A包含单词B,那么只计算单词A不计算单词B. dp[i][j]表示长度i(自动机上转移k步)后缀状态是自动机第j个结点的字符串的最大价值 dp[0][…

题目一串DNA最少需要修改几个基因使其不包含一些致病DNA片段. 这道题应该是AC自动机+DP的入门题了,有POJ2778基础不难写出来. dp[i][j]表示原DNA前i位(在AC自动机上转移i步)且后缀状态为AC自动机结点j的最少需要修改的基因数 转移我为人人型,从dp[i][j]向ATCG四个方向转移到dp[i+1][j'],如果结点被标记包含致病基因就不能转移. #include<cstdio> #include<cstring> #include<queue>…

题目:给出n个串,问最多能够选出多少个串,使得前面串是后面串的子串(按照输入顺序) 分析: 其实这题是这题SPOJ 7758. Growing Strings AC自动机DP的进阶版本,主题思想差不多. 对于这题来说,需要离线操作.dp转移也是很显然. 但是由于数据比较大,所以普通的沿着fail指针往上走,逐步更新答案会TLE. 考虑把fail指针反向,由于ac自动机的每个节点均有唯一的fail指针,若是沿着fail指针往上走,显然都会走到root,所以反向之后显然是一棵树,不妨称之为fail树…