[二进制枚举+LCS]Card Hand Sorting 题目描述 When dealt cards in the card game Plump it is a good idea to start by sorting the cards in hand by suit and rank. The different suits should be grouped and the ranks should be sorted within each suit. But the order o…

这个题其实由于只有4种花色的,那么每种花色排列的顺序,也不过是4!种,然后对于每种花色内部到底是升序还是降序,其实也可以直接暴力,一共也就4!*2^4种情况,然后直接进行排序就可以了,但是我们如何计算需要移动的位置呢???我们这样考虑,我们由于要保证内部有序,那么最后一定是一个升序或者降序,那么插入一张牌,实际上是相当改变内部相对位置,那么考虑无序的,我们肯定是找到最长的递增子序列,那么他们一定是不用互相移动的,而其他的肯定是要移动的,因为他们不满足前后的顺序,并且他们是肯定是要移动的,直接二进…

时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 Lara Croft, the fiercely independent daughter of a missing adventurer, must push herself beyond her limits when she discovers the island where her father disappeared. In this mysterious island, Lara finds a tomb…

任意门:http://hihocoder.com/problemset/problem/1829 Tomb Raider 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 Lara Croft, the fiercely independent daughter of a missing adventurer, must push herself beyond her limits when she discovers the island where her fath…

Lara Croft, the fiercely independent daughter of a missing adventurer, must push herself beyond her limits when she discovers the island where her father disappeared. In this mysterious island, Lara finds a tomb with a very heavy door. To open the do…

题意:平面上有n个点(1<=N<=1000),你的任务是让所有n个点连通,为此, 你可以新建一些边,费用等于两个端点的欧几里得距离的平方.另外还有q(0<=q<=8)个套餐(数量小,可枚举),可以购买,如果你购买了第i个套餐,该套餐 中的所有结点将变得相互连通,第i个套餐的花费为ci. 分析:按照刘汝佳的思路做的.首先求一次本身的最小生成树值,然后枚举购买的套餐(二进制枚举),每次购买了之后,将其权值设为0,并且加进最小生成树. #include<cstdio> #in…

B. New Year and Old Property time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The year 2015 is almost over. Limak is a little polar bear. He has recently learnt about the binary system. He…

题目链接:http://poj.org/problem?id=2784 Buy or Build Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 1528   Accepted: 592 Description World Wide Networks (WWN) is a leading company that operates large telecommunication networks. WWN would li…

题意:给出n头牛的得病的种类情况,一共有m种病,要求找出最多有K种病的牛的数目: 思路:二进制枚举(得病处为1,否则为0,比如得了2 1两种病,代号就是011(十进制就是3)),首先枚举出1的个数等于k的二进制数,然后跟所有的牛的代号一一比较,符合的   +1,找出其中和最大的:就是转换2进制麻烦,用位运算就好实现了,但是位运算不是很明白含义,明白了再补充: 知识点: 3 & 2 = 2,相同为1,不同为0, 011 & 010 = 010:(怎么利用的这个特点不明白),在计算机网络中也学…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3118 思路:题目要求是去掉最少的边使得图中不存在路径长度为奇数的环,这个问题等价于在图中去掉若干条边,使得这个图成为二分图.注意到n不是很大,于是我们可以想到二进制枚举,枚举每条边的两个顶点是否在同一个集合中,若是,则删除这条边. #include<iostream> #include<cstdio> #include<cstring> #include<algori…