P2862 [USACO06JAN]把牛Corral the Cows 题目描述 Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corral be square and that the corral contain at least C (1 <= C <= 500) clover fields for afternoon treats. The co…

P2862 [USACO06JAN]把牛Corral the Cows 题目描述 Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corral be square and that the corral contain at least C (1 <= C <= 500) clover fields for afternoon treats. The co…

P2862 [USACO06JAN]把牛Corral the Cows 题目描述 Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corral be square and that the corral contain at least C (1 <= C <= 500) clover fields for afternoon treats. The co…

P2862 [USACO06JAN]把牛Corral the Cows 题目描述 Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corral be square and that the corral contain at least C (1 <= C <= 500) clover fields for afternoon treats. The co…

题目描述 约翰打算建一个围栏来圈养他的奶牛.作为最挑剔的兽类,奶牛们要求这个围栏必须是正方 形的,而且围栏里至少要有C< 500)个草场,来供应她们的午餐. 约翰的土地上共有C<=N<=500)个草场,每个草场在一块1x1的方格内,而且这个方格的 坐标不会超过10000.有时候,会有多个草场在同一个方格内,那他们的坐标就会相同. 告诉约翰,最小的围栏的边长是多少? 输入输出格式 输入格式: Line 1: Two space-separated integers: C and N Lin…

传送门 可以二分边长 然后另开两个数组,把x从小到大排序,把y从小到大排序 枚举x,可以得到正方形的长 枚举y,看看从这个y开始,往上能够到达多少个点,可以用类似队列来搞 其实发现算法的本质之后,x可以不用从小到大排序 #include <cstdio> #include <iostream> #include <algorithm> #define N 1001 #define max(x, y) ((x) > (y) ? (x) : (y)) int c, n…

洛谷P1522 [USACO2.4]牛的旅行 Cow Tours 题意: 给出一些牧区的坐标,以及一个用邻接矩阵表示的牧区之间图.如果两个牧区之间有路存在那么这条路的长度就是两个牧区之间的欧几里得距离. 对于一个联通块,称之为一个牧场,也就是说一个牧场内任意一个牧区都可以到达该牧场内的任意的另外一个牧区. 对于一个牧场,它的直径是这个联通块内最短路的最大值. 现在让你在恰当地选择两个牧场,在这两个牧场中各自选一个牧区,在这两个牧区之间建路,要求建路之后所有牧场中最大的直径最小.这里其实如果产生了…

P1821 [USACO07FEB]银牛派对Silver Cow Party 题目描述 One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads co…

本来分好组之后,就确定好了每个人要学什么,我去学数据结构啊. 因为前一段时间遇到一道题是用Lca写的,不会,就去学. 然后发现Lca分为在线算法和离线算法,在线算法有含RMQ的ST算法,前面的博客也写了.离线算法是基于DFS的Tarjan算法. 然后就打算去学一下Tarjan,因为以前也看过但是没看完,就打算学一下,因为Tarjan算法是图论的内容,然后就让图论选手教了我一下大环套小环的怎么推,然后就尴尬了. 我是学数据结构的,没有要去抢图论的内容学... 我也看了线段树了啊. 学完这个Tarj…

传送门 题目大意:形成一个环的牛可以跳舞,几个环连在一起是个小组,求几个小组. 题解:tarjian缩点后,求缩的点包含的原来的点数大于1的个数. 代码: #include<iostream> #include<cstdio> #include<cstring> #define maxn 10009 using namespace std; int n,m,sumedge,top,sumclr,tim,ans; int Stack[maxn],instack[maxn]…