Diamond Collector 题目描述 Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare time! She has collected N diamonds (N≤50,000) of varying sizes, and she wants to arrange some of them in a pair of display case…

P3143 [USACO16OPEN]钻石收藏家Diamond Collector 题目描述 Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare time! She has collected \(N\) diamonds (\(N \leq 50,000\) of varying sizes, and she wants to arrange so…

问题 I: Diamond Collector 时间限制: 1 Sec  内存限制: 64 MB提交: 22  解决: 7[提交][状态][讨论版] 题目描述 Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare time! She has collected N diamonds (N≤50,000) of varying sizes, and sh…

4582: [Usaco2016 Open]Diamond Collector Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 204  Solved: 136[Submit][Status][Discuss] Description Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare  time! S…

P3143 [USACO16OPEN]钻石收藏家Diamond Collector 题目描述 Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare time! She has collected NN diamonds (N \leq 50,000N≤50,000) of varying sizes, and she wants to arrange…

bzoj4582[Usaco2016 Open]Diamond Collector 题意: n个钻石,每个都有一个大小,现在将其装进2个盒子里,每个盒子里的钻石最大的与最小的大小不能超过k,问最多能装多少个.n最大50000. 题解: 我真傻,真的~首先对大小排序,然后找以i为左端点的可装区间,这个操作两个指针就可以搞,我却以为要二分查找.预处理完了,因为不交错的区间肯定比交错的区间优,所以从n到1递推一下从n到i最大的区间大小是多少,然后枚举每个区间,找到当前区间大小加上右端点+1到n中最大的…

http://www.lydsy.com/JudgeOnline/problem.php?id=4582 排好序后用两个指针直接\(O(n)\)扫,貌似这个东西学名"two pointers"? #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 50003; int pre[N]…

Descrirption 给你一个长度为 \(n\) 的序列,求将它分成两个序列后最多个数,每个序列最大值最小值不能超过 \(k\) Sol 二分+DP. 排一下序,找出以这个点结尾和开始的位置. 这个玩意可以二分也可以用单调队列,随便搞啊... 然后统计答案就是枚举第二个序列的起点,然后往后扫的时候统计一下,第一个序列的最大长度就可以了. Code /************************************************************** Problem:…

算是一道dp 首先,排序好每一个架子上都是一段区间,然后只需要统计每个点向左向右最长延伸的区间. 所以我们预处理出每个点以左.以右最大能延伸的长度(最多能选几个差值不超过k的) 然后枚举每个点作为断点,sum取max即可 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ; int n,k,a[N],sum_l[N],sum_r[N]; int read(){ ; ch…

由于相差不超过k才可以放在一起,要判断不超过k这个条件,显然我们需要排序 首先我们需要一个f数组,f[i]意义看代码开头注释, 假设我们可以选择的某一个区间是a[l]~a[r](已排序且最优(最长的意思)),那么这个区间要符合这样一个性质: a[r]-a[l]<=k 移项一下, a[r]<=k+a[l](r要尽可能大) 而a[l]中的l我们是枚举得到的,因此我们只需要求出对应的r即可 我们观察移项后的式子可以发现,我们要求的r其实是数列中小于等于k+a[l]的最大值 那么这显然是符合可二分性的…