3018: [Usaco2012 Nov]Distant Pastures Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 43  Solved: 20[Submit][Status][Discuss] Description Farmer John's farm is made up of an N x N grid of pastures, where each pasture contains one of two different type…

3016: [Usaco2012 Nov]Clumsy Cows Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 71  Solved: 52[Submit][Status] Description Bessie the cow is trying to type a balanced string of parentheses into her new laptop, but she is sufficiently clumsy (due to h…

3016: [Usaco2012 Nov]Clumsy Cows Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 91  Solved: 69[Submit][Status][Discuss] Description Bessie the cow is trying to type a balanced string of parentheses into her new laptop, but she is sufficiently clumsy…

http://www.lydsy.com/JudgeOnline/problem.php?id=3016 之前yy了一个贪心,,,但是错了,,就是枚举前后对应的字符(前面第i个和后面第i个)然后相同答案就+1,否则不操作.. QAQ 然后看了题解...神.. 首先序列肯定是偶数个,然后..一定有n/2个‘(’,和n/2个‘)’ 那么左边的一定都是‘(’,对于每一个‘(’,右边的一定有一个‘)’,所以我们想到差分,当右边多了‘)’,那么就要累计答案+1,维护一个sum,如果是‘(’就+1,‘)’就…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=3016 题意: 给你一个括号序列,问你至少修改多少个括号,才能使这个括号序列合法. 题解: 贪心. cnt表示当前已经攒了多少个左括号. 从左往右枚举每一个括号: (1)如果为左括号,则cnt++. (2)如果为右括号,且cnt > 0,则cnt--.表示消去了一个左括号. (3)如果为右括号,且cnt <= 0,则cnt++,ans++.因为此时一定要改. 最后ans还要加上cnt/2…

Description [Brian Dean, 2012] Farmer John's cows are all of a very peculiar breed known for its distinctive appearance -- each cow is marked with a giant spot on its hide in the shape of a parenthesis (depending on the direction the cow is facing, t…

bzoj上的usaco题目还是很好的(我被虐的很惨. 有必要总结整理一下. 1592: [Usaco2008 Feb]Making the Grade 路面修整 一开始没有想到离散化.然后离散化之后就很好做了.F[I,j]表示第i个点,高度>=j或<=j,f[I,j]=min(f[i-1,j]+abs(b[j]-a[i]),f[I,j-1]) 1593: [Usaco2008 Feb]Hotel 旅馆 线段树 ★1594: [Usaco2008 Jan]猜数游戏 二分答案然后写线段树维护 15…

Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…

最近刷银组刷得好欢快，好像都是水题，在这里吧他们都记录一下吧（都是水题大家一定是道道都虐的把= =）几道比较神奇的题到时再列出来单独讲一下吧= =（其实我会说是BZOJ蹦了无聊再来写的么 = =） [Usaco2004 Dec]Bad Cowtractors牛的报复  很明显是最大生成树了吧，跟最小生成树一样做就行了 = =（排序时按从大到小的顺序排） [Usaco2004 Dec]Cleaning Shifts安排值班  贪心，按开始时间从小到大排，然后对于已经安排的时间，取在此时间开始的最晚…

https://files.cnblogs.com/files/Winniechen/usaco2012-2013.pdf 做的不是很好,还请见谅! 如果有什么疑问,可以QQ上找我. QQ号:1967199892 附上代码: BZOJ3010: [Usaco2012 Dec]Gangs of Istanbull #include <cstdio> #include <cmath> #include <algorithm> #include <iostream>…