SuperBull bzoj-3943 Usaco-2015 Feb 题目大意：贝西和她的朋友们在参加一年一度的“犇”(足)球锦标赛。FJ的任务是让这场锦标赛尽可能地好看。一共有N支球队参加这场比赛，每支球队都有一个特有的取值在1-230-1之间...
SuperBull bzoj-3943 Usaco-2015 Feb 题目大意：贝西和她的朋友们在参加一年一度的“犇”(足)球锦标赛。FJ的任务是让这场锦标赛尽可能地好看。一共有N支球队参加这场比赛，每支球队都有一个特有的取值在1-230-1之间...
题目描述 Bessie and her friends are playing hoofball in the annual Superbull championship, and Farmer John isin charge of making the tournament as exciting as possible....= N <= 20...
Description Bessie and her friends are playing hoofball in the annual Superbull championship, and Farmer John is in charge of making the tournament as exciting as possible. A total of N (1 ...
有任何问题欢迎留言或私聊 题目链接：BZOJ4390 题目： Description Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), ...
Just like humans enjoy playing the game of Hopscotch, Farmer John's cows have invented a variant of the game for themselves to play. Being played by clumsy animals weighing nearly a ton, Cow Hopscot.....
[题目链接] https://www.lydsy.com/JudgeOnline/problem.php?id=4392 [算法] 线段树 时间复杂度 ： O(MlogN) [代码] #include<bits/stdc++.h> using namespace std;...#defi...
Description Bessie the cow is a huge fan of card games, which is quite surprising, given her lack of opposable thumbs. Unfortunately, none of the other cows in the herd are good opponents....
41.USACO历年青铜组真题解析 | 2023年12月Cowntact Tracing 2-程序员宅基地。22.USACO历年青铜组真题解析 | 2021年12月Air Cownditioning-程序员宅基地。3.USACO历年青铜组真题解析 | 2022年12月Feeding the Cows-程序员宅基地。...
原题链接https://www.lydsy.com/JudgeOnline/problem.php?id=4397 用线段树维护区间和即可。时间复杂度\(O((N+Q)logN)\)。 #include<iostream> #include<cstring> #include<...using name...
第一题裸四方暴力，一直在想怎么用数据结构优化，期间想到了主席树套二维线段树，然而会爆空间，但是只要从上到下从左到右便可以只要主席树了，然后动态开点就是O(n2logn)O(n^2logn)的空间，还是对主席树不熟练啊，...
3943: [Usaco2015 Feb]SuperBull Time Limit:10 SecMemory Limit:128 MBSubmit:300Solved:185 Description Bessie and her friends are playing hoofball in the annual Superbull championsh...
广告：#include int main() { puts("转载请注明出处[vmurder]谢谢"); puts("网址：blog.csdn.net/vmurder/article/details/44065939"); }题意：PoPoQQQPoPoQQQ站在原点上向y轴正半轴看，然后有一群羊驼从他眼前...
Problem 2. Fruit Feast 很简单的智商题（因为碰巧脑出来了所以简单一，一 原题： Bessie has broken into Farmer John's house again! She has discovered a pile of lemons and a pile of oranges in the ...
画廊秀坊霓裳舞，小桥流水叶娉婷
Description Just like humans enjoy playing the game of Hopscotch, Farmer John’s cows have invented a variant of the game for themselves to play. Being played by...
题目描述 给定一棵有N个点的树，所有节点的权值都为0。 有K次操作，每次指定两个点s,t，将s到t路径上所有点的权值都加一。 请输出K次操作完毕后权值最大的那个点的权值。 输入格式 The first line of the input ...
题意比较难理解，就是给你n个点的树，然后给你m个修改操作，每一次修改包括一个点对(x, y)，意味着将x到y所有的点权值加一，最后问你整个树上的点权最大是多少。 比较裸的树链剖分了，感谢Haild的讲解。...
BZOJ3887 [Usaco2015 Jan] Grass Cownoisseur Description In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all ov...
广搜 保证不重复进队 然后多搜几遍直到不能更新答案为止 #include #include #include #include #include #include #include #include #include #include #define T 43333 using namespace std;... whi
Time Limit:10 SecMemory Limit:128 MBSubmit:391Solved:183 Description Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty ...
USACO 2020年12月铜组
应江老师的要求，我将官方结题报告中的三、四题的java代码翻译成了C++。顺便还将注释翻译成了中文。但是还没有检查。3、注：该题检查已过。# include # include using namespace std;const int MAXN = 4001;...
对kmp要有更深层次的理解呢
Description Farmer John's N cows, conveniently numbered 1…N, are all standing in a row (they seem to do so often that it now takes very little prompting from Farmer John to line them up)....
【USACO 2015 Feb Gold】检查问题描述 FJ把杂志上所有的文章摘抄了下来并把它变成了一个长度不超过10^5的字符串S。他有一个包含n个单词的列表，列表里的n个单词记为t_1…t_N。他希望从S中删除这些单词。 FJ每次在...
bzoj3940[Usaco2015 Feb]Censoring 题意: 有一个S串和一大堆T串，不断地在S串里找最早出现的T串，然后将其删除。S串≤100000，T串总长度≤100000。 题解： 对所有T串建AC自动机，然后同bzoj3942。注意，本题的AC...
bzoj3943[Usaco2015 Feb]SuperBull 题意： n头牛进行锦标赛，每场比赛的好看程度是两头牛的编号异或和，并总有一方被淘汰。求安排比赛（可以决定比赛胜负）可以得到的最大总好看程度是多少。n≤2000 题解： 先求...
bzoj3942[Usaco2015 Feb]Censoring 题意： 有一个S串和一个T串，不断地在S串里匹配T串，然后将其删除。S串、T串长度≤1000000。 题解： 用1、2两个栈，每次将S串的当前字符压入1栈，当前匹配到T串的位置压入2栈...
题目描述传送门题解和BZOJ3942是差不多的。 搞一个栈，然后边匹配边压栈，匹配不了就跳到失配，匹配成功就暴力弹栈。 需要注意的一点是，弹栈之后要将当前匹配的节点恢复到栈顶元素的位置，然后从下一个元素开始...
题目描述传送门题解首先kmp求出来失配函数，然后暴力匹配。 如果当前位可以匹配，那么将其压入栈中；如果不能匹配，蹦到它的失配开始匹配；如果有一个完整的子串被压入栈中了，暴力将这个子串弹出。...