我一开始写了个状压dp..然后没有滚动就MLE了... 其实这道题直接暴力就行了... 2^15枚举每个状态, 然后检查每头牛是否能被选中, 这样是O( 2^15*1000 ), 也是和dp一样的时间复杂度....有点贪心的feel ------------------------------------------------------------------------------ #include<cstdio> #include<algorithm> #include&l…

1688: [Usaco2005 Open]Disease Manangement 疾病管理 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 413  Solved: 275[Submit][Status][Discuss] Description Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John woul…

http://www.lydsy.com/JudgeOnline/problem.php?id=1688 很水的状压.. 提交了很多次优化的,但是还是100msT_T #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue>…

Description Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K (1 <= K <= D) diff…

#include <bits/stdc++.h> #define setIO(s) freopen(s".in","r",stdin) #define maxn 100000 using namespace std; int F[1 << 16], s[1 << 16], v[1 << 16], f[1 << 16]; int n, D, k, tot = 0, ans = 0; int main() {…

[BZOJ1688][Usaco2005 Open]Disease Manangement 疾病管理 Description Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the mil…

vs(i)表示患i这种疾病的牛的集合. f(S)表示S集合的病被多少头牛患了. 枚举不在S中的疾病i,把除了i和S之外的所有病的牛集合记作St. f(S|i)=max{f(S)+((St|vs(i))^St)中牛的数量} #include<cstdio> #include<bitset> #include<algorithm> using namespace std; bitset<1000>vs[15],t,t2; int n,m,K,ans,f[1<…

思路:状压dp,枚举疾病的集合,然后判断一下可行性即可. #include<bits/stdc++.h> using namespace std; #define maxs 400000 #define maxn 1900 int n,d,k; int a[maxn],f[maxs],num[maxs]; int main(){ scanf("%d%d%d",&n,&d,&k); ;i<(<<d);i++) num[i]=num[i…

分析: 这个题的状压DP还是比较裸的,考虑将疾病状压,得到DP方程:F[S]为疾病状态为S时的最多奶牛数量,F[S]=max{f[s]+1}; 记得预处理出每个状态下疾病数是多少... 附上代码: #include <cstdio> #include <algorithm> #include <cmath> #include <cstdlib> #include <cstring> #include <queue> #include…

题目描述 Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K (1 <= K <= D) different d…

题目描述 Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K (1 <= K <= D) different d…

Disease Manangement 疾病管理   Description Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than…

题目描述 Alas! \(A\) set of \(D (1 <= D <= 15)\) diseases (numbered \(1..D\)) is rshning through the farm. Farmer John would like to milk as many of his N \((1 <= N <= 1,000)\) cows as possible. If the milked cows carry more than K \((1 <= K &l…

Disease Manangement Q - 枚举子集 Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u   Description Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many…

时光匆匆,不知不觉就到了第十二周.——第一次迭代都完成了,最终迭代还会远吗? 一.第一次迭代的设想和目标: 1.  我们的软件要解决什么问题?是否定义得很清楚?是否对典型用户和典型场景有清晰的描述? 软件要解决的问题:针对慢阻肺病人的后期康复,减少医患沟通负担,给予病人一个更好的平台去问医.监督个人康复进程. 典型用户:慢阻肺病人和治疗医生. 典型场景:家庭治疗. 2. 我们达到目标了么(原计划的功能做到了几个? 按照原计划交付时间交付了么? 原计划达到的用户数量达到了么?)? 原计划:实现登陆…

一.项目介绍: 先上图: 这是我们小组开发项目的系统构成图. 本项目研发面向家庭/社区,对稳定期中慢阻肺患者病情和环境全方位监测.简单病情趋势分析和患者行为干预,并且提供患者与家庭医生和专家沟通渠道的慢阻肺患者康复综合护理系统. (2018-12-07:补充: 本app用于慢阻肺足不出户家庭护理.分为患者端和医生端. 患者和医生可以实现绑定. 已绑定的患者和医生可以实现在线沟通:问诊.随访...... 医生可以根据患者的情况制定护理计划并发送给患者.患者根据医生制定的具体的护理计划,定时服药.使…

问题描述 BZOJ1688 题解 背包,在转移过程中使用状压. \(\mathrm{Code}\) #include<bits/stdc++.h> using namespace std; template <typename Tp> void read(Tp &x){ x=0;char ch=1;int fh; while(ch!='-'&&(ch>'9'||ch<'0')) ch=getchar(); if(ch=='-') ch=getch…

Description Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K (1 <= K <= D) diff…

大神们都在刷usaco,我也来水一水 1606: [Usaco2008 Dec]Hay For Sale 购买干草   裸背包 1607: [Usaco2008 Dec]Patting Heads 轻拍牛头 神转化,筛法 1609: [Usaco2008 Feb]Eating Together麻烦的聚餐  LIS 1610: [Usaco2008 Feb]Line连线游戏 排序 1611: [Usaco2008 Feb]Meteor Shower流星雨  BFS 1612: [Usaco2008…

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…

由于我太菜,决定按照AC人数从小到大慢慢做. BZOJ开了权限号真的快了好多诶~ 29/50 1606: [Usaco2008 Dec]Hay For Sale 购买干草 背包dp 1610: [Usaco2008 Feb]Line连线游戏 把斜率算出来去重 *1607: [Usaco2008 Dec]Patting Heads 轻拍牛头 题意不清,题意.筛法,详见hzwer博客 1609: [Usaco2008 Feb]Eating Together麻烦的聚餐 正反都搞一遍LIS,然后结果就是…

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…

Disease Manangement 疾病管理 Description Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K…

因为篇幅太长翻着麻烦,计划把DP拆成几个小专题,这里原文只留下状压,其他请至后续博文. 状态压缩优化 所谓状态压缩,就是将原本需要很多很多维来描述,甚至暴力根本描述不清的状态压缩成一维来描述. 时间复杂度一般为\(O(2^n\cdot n^2)\)的形式 (ZZ并不太会算复杂度,如果博客中复杂度有错误,请指出并尽情嘲讽我,谢谢!) 眼界极窄的ZZ之前只是听说过这个名字--先感谢Lrefrain学长把这个东西介绍给我orz 使用状态压缩优化的常见情景: 这个数据范围怎么有一维出奇的小啊? 互不侵犯…

[BZOJ1606][Usaco2008 Dec]Hay For Sale 裸01背包 ;i<=n;i++) for(int j=m;j>=a[i];j--) f[j]=max(f[j],f[j-a[i]]+a[i]); [BZOJ1607][Usaco2008 Dec]轻拍牛头 暴力 [BZOJ1609][Usaco2008 Feb]麻烦的聚餐 要求最长上升子序列或最长下降子序列的长度. 最基础的有dp算法 ;i<=n;i++) ;j<=;j++){ ;k<=j;k++)f…

T1 疾病管理 裸得不能再裸的状压dp 不过数据范围骗人 考试时k==0的点没过 我也很无奈呀qwq #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<string> using namespace std; ],f[<<]; int pd(int x){ ; while…

新概念二 Lesson95 词汇 ①get a shock 吓了一跳,得到一个惊喜 例:his wife got a shock get into a such mess 这么不幸搞得一片狼籍弄得这样一塌糊涂 例:How did your clothes get into a such mess? ③set on fire 放火 例:University set the Embassy on fire this morning 私exclaim V.大叫,呼喊 ⑤break out 爆发出 Ac…

短视频APP——昙花一现还是发展趋势? 在这个互联网与科技并行且飞速发展的时代,各种app不断涌入市场,其中短视频app便是一个典型,美拍,就成功入围2014年十大最火app.而短视频app也势必要成为发展趋势而绝非昙花一现,接下来就让我们一起分析一下它之所以能够引领时代潮流的种种原因. (一)时代背景 1.当前网络上的传播媒体十分多样,诸如微信,微博,QQ等都能够随时随地让人们分享自己的幸福,展示自己的风采,这样一来,便为短视频app提供了更广阔的交流发布空间. 2.短视频app的种类多种多样…

实现原理: 创建多个div,div之间通过css实现层叠,每个div放置当前表格的克隆.例如:需要行冻结时,创建存放冻结行表格的div,通过设置z-index属性和position属性,让冻结行表格在数据表格的上层.同理,需要列冻结时,创建存放冻结列表格的div,并放置在数据表格的上层.如果需要行列都冻结时,则除了创建冻结行.冻结列表格的div,还需要创建左上角的固定行列表格的div,并放置在所有div的最上层. 处理表格的滚动事件,在表格横向或者纵向滚动时,同时让相应的冻结行和冻结列也同步滚动…

开始更新咯 DP专题[题目来源BZOJ] 一.树形DP 1.bzoj2286消耗战 题解:因为是树形结构,一个点与根节点不联通,删一条边即可, 于是我们就可以简化这棵树,把有用的信息建立一颗虚树,然后开始DP即可 /* 思路: */ #include<algorithm> #include<cstdio> #include<cmath> #include<iostream> #include<cstring> #define ll long lo…