Cellular automata are mathematical idealizations of physical systems in which both space and time are discrete, and the physical quantities take on a nite set of discrete values. A cellular automaton consists of a lattice (or array), usually in nite,…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5977 Description When God made the first man, he put him on a beautiful garden, the Garden of Eden. Here Adam lived with all animals. God gave Adam eternal life. But Adam was lonely in the garden, s…

题目链接: Garden of Eden Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 210    Accepted Submission(s): 75 Problem Description When God made the first man, he put him on a beautiful garden, the G…

Problem Description When God made the first man, he put him on a beautiful garden, the Garden of Eden. Here Adam lived with all animals. God gave Adam eternal life. But Adam was lonely in the garden, so God made Eve. When Adam was asleep one night, G…

Garden of Eden Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description When God made the first man, he put him on a beautiful garden, the Garden of Eden. Here Adam lived with all animals. God gave Ad…

都不好意思写题解了 跑了4000多ms 纪念下自己A的第二题 (我还有一道freetour II wa20多发没A...呜呜呜 #include<bits/stdc++.h> using namespace std; #define sz(X) ((int)X.size()) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define index Index typedef long long ll; const…

题解: 路径统计比较容易想到点分治和dp dp的话是f[i][j]表示以i为根,取了i,颜色数状态为j的方案数 但是转移这里如果暴力转移就是$(2^k)^2$了 于是用FWT优化集合或 另外http://www.cnblogs.com/sclbgw7/p/9508235.html给出了一种技巧优化空间 就是我们优先处理重儿子,这样子我们上面记录的状态一定都是连着轻边的 而由树链剖分的复杂度证明我们可以知道一条路径上轻边最多只有log条 为什么呢,因为重儿子肯定比轻儿子大,所以至少翻倍…

题目大意:求有所有颜色的路径数. 题目分析:参考codeforces997C,先利用基的FMT的性质在$O(2^k)$做FMT,再利用只还原一位的特点在$O(2^k)$还原,不知道为什么网上都要点分治. 代码: #include<bits/stdc++.h> #define R register using namespace std; ; int n,k,a[maxn],fa[maxn],cnt[maxn],head[maxn]; struct edge{int to,nxt;}edges[…

CGZ大佬提醒我,我要是再不更博客可就连一月一更的频率也没有了... emmm,正好做了一道有点意思的题,就拿出来充数吧=.= 题意 一棵树,有 $ n (n\leq50000) $ 个节点,每个点都有一个颜色,共有 $ k(k\leq10) $ 种颜色,问有多少条路径可以遍历到所有 $ k $ 种颜色?(一条路径交换起点终点就算两条哦) 做法 事实证明,连我都能不看题解想出来的题果然都是水题qwq 我是从CJ的xzyxzy大佬上的博客上看到这道题的,所以就理所当然用FWT做了...然后才发现网…

题意:给一棵节点数为n,节点种类为k的无根树,问其中有多少种不同的简单路径,可以满足路径上经过所有k种类型的点? 析:对于路径,就是两类,第一种情况,就是跨过根结点,第二种是不跨过根结点,分别讨论就好,由于结点比较大,所以采用分治来进行处理,优先选取重点作为划分的依据. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string>…