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…

pid=2825">http://acm.hdu.edu.cn/showproblem.php? pid=2825 Wireless Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4560    Accepted Submission(s): 1381 Problem Description Liyuan li…

Wireless Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4022    Accepted Submission(s): 1196 Problem Description Liyuan lives in a old apartment. One day, he suddenly found that there…

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…

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…

给m个单词,由这m个单词组成的一个新单词(两个单词可以重叠包含)长度为n,且新单词中包含的基本单词数目不少于k个.问这样的新单词共有多少个? m很小,用二进制表示新单词中包含基本单词的情况. 用m个单词建立AC自动机,可以求出所有单词之间相互包含的情况,AC自动机的后缀特性(每个结点的失配边指向新结点,新节点到trie树根的字符串是当前节点字符串的后缀). 动态规划: f(i, j, k):长度为i的单词,在trie树中第j个结点处,包含基本单词的情况为k(二进制),的总方案数. 转移方程: 在…

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…

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…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=2825 题意:给一些字符串,构造出长度为n的字符串,它至少包含k个所给字符串,求能构造出的个数. 题解: 对end[]节点标记数组进行改动,用二进制下第几位表示即为包含第几个给定子串: 状态转移方程为:dp[i+1][x][k|end[x]]=(dp[i+1][x][k|end[x]]+dp[i][j][k])%mod:(没懂为什么要用' | ' ,等搞明白了在更新吧,,) dp[i][j][l] 表…

题意:m个密码串,问你长度为n的至少含有k个不同密码串的密码有几个 思路:状压一下,在build的时候处理fail的时候要用 | 把所有的后缀都加上. 代码: #include<cmath> #include<set> #include<map> #include<queue> #include<cstdio> #include<vector> #include<cstring> #include <iostream…