AC日记——Keywords Search hdu 2222】的更多相关文章

2222 思路: ac自动机模板题: 代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 500005 struct TreeNodeType { int count; TreeNodeType *fail; TreeNodeType *next[]; TreeNodeType()…
Keywords Search Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 77903    Accepted Submission(s): 27032 Problem Description In the modern time, Search engine came into the life of everybody lik…
In the modern time, Search engine came into the life of everybody like Google, Baidu, etc. Wiskey also wants to bring this feature to his image retrieval system. Every image have a long description, when users type some keywords to find the image, th…
题目大意:输入几个子串,然后输入一个母串,问在母串里面包含几个子串.   分析:刚学习的AC自动机,据说这是个最基础的模板题,所以也是用了最基本的写法来完成的,当然也借鉴了别人的代码思想,确实是个很神奇的东西,如果不懂KMP的话,最好先学学KMP再来学这个,会理解的更深刻一些.   代码如下: ======================================================================================================…
求一个字符串上有多少个匹配的单词 看着卿学姐的板子写的 指针形式: #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include…
题目网址:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=110773#problem/A Description In the modern time, Search engine came into the life of everybody like Google, Baidu, etc. Wiskey also wants to bring this feature to his image retrieval system. …
Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 24928    Accepted Submission(s): 10551 Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1],…
统计难题 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131070/65535 K (Java/Others)Total Submission(s): 36637    Accepted Submission(s): 13622 Problem Description Ignatius最近遇到一个难题,老师交给他很多单词(只有小写字母组成,不会有重复的单词出现),现在老师要他统计出以某个字符串为前缀的单词数量(单词本身也是自己的…
2896 思路: 好题: 代码: #include <queue> #include <cstdio> #include <cstring> using namespace std; #define maxn 100001 ][],tag[],fail[],n,tot,m,ans; char ch[maxn]; ],]; queue<int>que; int main() { int i,j,len,now,temp; scanf("%d"…
Paint Pearls 思路: 离散化+dp+剪枝: dp是个n方的做法: 重要就在剪枝: 如果一个长度为n的区间,有大于根号n种颜色,还不如一个一个涂: 来,上代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 50005 int n,ai[maxn],dp[maxn],ls[…