USACO Mooo Moo
洛谷 P2214 [USACO14MAR]哞哞哞Mooo Moo
JDOJ 2416: USACO 2014 Mar Silver 3.Mooo Moo
Description
Problem 3: Mooo Moo [silver] [Brian Dean, 2014]
Farmer John has completely forgotten how many cows he owns! He is too
embarrassed to go to his fields to count the cows, since he doesn't want
the cows to realize his mental lapse. Instead, he decides to count his
cows secretly by planting microphones in the fields in which his cows tend
to gather, figuring that he can determine the number of cows from the total
volume of all the mooing he hears.
FJ's N fields (1 <= N <= 100) are all arranged in a line along a long
straight road. Each field might contain several types of cows; FJ
owns cows that come from B different breeds (1 <= B <= 20), and a cow
of breed i moos at a volume of V(i) (1 <= V(i) <= 100). Moreover,
there is a strong wind blowing down the road, which carries the sound
of mooing in one direction from left to right: if the volume of mooing
in some field is X, then in the next field this will contribute X-1 to
the total mooing volume (and X-2 in the field after that, etc.).
Otherwise stated, the mooing volume in a field is the sum of the
contribution due to cows in that field, plus X-1, where X is the total
mooing volume in the preceding field.
Given the volume of mooing that FJ records in each field, please compute
the minimum possible number of cows FJ might own.
The volume FJ records in any field is at most 100,000.
Input
* Line 1: The integers N and B.
* Lines 2..1+B: Line i+1 contains the integer V(i).
* Lines 2+B..1+B+N: Line 1+B+i contains the total volume of all mooing
in field i.
Output
* Line 1: The minimum number of cows owned by FJ, or -1 if there is no
configuration of cows consistent with the input.
Sample Input
5 2 5 7 0 17 16 20 19
Sample Output
4
HINT
INPUT DETAILS:
FJ owns 5 fields, with mooing volumes 0,17,16,20,19. There are two breeds
of cows; the first moos at a volume of 5, and the other at a volume of 7.
OUTPUT DETAILS:
There are 2 cows of breed #1 and 1 cow of breed #2 in field 2, and there is
another cow of breed #1 in field 4.
题目翻译:
约翰忘记了他到底有多少头牛,他希望通过收集牛叫声的音量来计算牛的数量。
他的N (1 <= N <= 100)个农场分布在一条直线上,每个农场可能包含B (1 <= B <= 20)个品种的牛,一头品种i的牛的音量是V(i) ,(1 <= V(i) <= 100)。一阵大风将牛的叫声从左往右传递,如果某个农场的总音量是X,那么将传递X-1的音量到右边的下一个农场。另外,一个农场的总音量等于该农场的牛产生的音量加上从上一个农场传递过来的音量(即X-1)。任意一个农场的总音量不超过100000。
请计算出最少可能的牛的数量。
题解:
完全背包问题的一个小变形。
设置dp[i] 为音量为i时最少的奶牛数量,所以最后的答案就是所有牧场自己的音量的总和。
注意,是实际音量,不是实际音量。
所以在原有a数组保存农场总音量的同时,我们引入了b数组来保存这个牧场的单纯音量。
注意,这里统计b数组的时候,一定要加max(和0比较),否则会WA3个点。
然后用maxx统计最大值,这样初始化的时候可以节省一点点时间。
然后就是振奋人心的DP过程啦!套用完全背包的模板的时候注意要加一个判断。
这个判断就是一种优化,emm,怎么说呢》自己体会吧!。
AC CODE:
#include<cstdio>
#include<algorithm>
using namespace std;
const int INF=1e9;
int n,B,maxx=-1,ans=0;
int v[25],a[105],b[105],dp[100005];
int main()
{
scanf("%d%d",&n,&B);
for(int i=1;i<=B;i++)
scanf("%d",&v[i]);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
{
b[i]=a[i]-max(a[i-1]-1,0);
maxx=max(maxx,b[i]);
}
for(int i=1;i<=maxx;i++)
dp[i]=INF;
for(int i=1;i<=B;i++)
for(int j=v[i];j<=maxx;j++)
if(dp[j-v[i]]!=INF)
dp[j]=min(dp[j],dp[j-v[i]]+1);
for(int i=1;i<=n;i++)
{
if(dp[b[i]]==INF)
{
printf("-1");
return 0;
}
ans+=dp[b[i]];
}
printf("%d",ans);
return 0;
}
USACO Mooo Moo的更多相关文章
- 【题解】Luogu P2214 [USACO14MAR]哞哞哞Mooo Moo
P2214 [USACO14MAR]哞哞哞Mooo Moo 题目描述 Farmer John has completely forgotten how many cows he owns! He is ...
- (寒假集训)Mooo Moo (完全背包)
Mooo Moo 时间限制: 1 Sec 内存限制: 64 MB提交: 5 解决: 4[提交][状态][讨论版] 题目描述 Farmer John has completely forgotten ...
- P2214 [USACO14MAR]哞哞哞Mooo Moo
链接:Miku ---------------------- 这道题还是个背包 --------------------- 首先看一下声音的组成,对于每一个农场的声音,它是由两部分组成的 :上一个农场 ...
- USACO 2020 OPEN Silver Problem 3. The Moo Particle
题意: 解法: 首先给出在本题中连通和连通块的定义: 连通: 两个粒子a,b连通,当且仅当ax≤bx.ay≤by或者bx≤ax.by≤ay. 如图,A,B两粒子是连通的,而C.D不是. 可以看出,本题 ...
- usaco silver
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 ...
- Moo University - Financial Aid
Moo University - Financial Aid Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 6020 Accep ...
- Bzoj 1657: [Usaco2006 Mar]Mooo 奶牛的歌声 单调栈
1657: [Usaco2006 Mar]Mooo 奶牛的歌声 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 631 Solved: 445[Submi ...
- BZOJ1657: [Usaco2006 Mar]Mooo 奶牛的歌声
1657: [Usaco2006 Mar]Mooo 奶牛的歌声 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 489 Solved: 338[Submi ...
- poj 2010 Moo University - Financial Aid(优先队列(最小堆)+ 贪心 + 枚举)
Description Bessie noted that although humans have many universities they can attend, cows have none ...
随机推荐
- 解密httpclient,dbcp,jedis,c3p0,druid,okhttp都在使用的连接池技术
最近在连接池上面栽了个跟头(参见这里),引起我对池技术的强烈关注,这几天总结了一下很多场景都会使用的池技术: 池概念 pool,中文翻译为水池,但是在英文中,还有一种解释是 an organizati ...
- Shell脚本——添加和删除用户
写一个脚本admin_user.sh,其用法格式为: admin_user.sh --add USERLIST --del USERLIST -v|--verbose -h|--help 其中, -h ...
- 使用thanos管理Prometheus持久化数据
关于thanos的介绍可以参考这篇官方博客的翻译文档,本文不作部署操作介绍.下图是thanos的官方架构图,主要有5个组件: Query:可以近似看作是Prometheus的实现,用于采集其他组件的数 ...
- rem与em的使用和区别
区别是:浏览器根据谁来转化成px值. 当使用rem单位,转换为像素大小取决于根元素的字体大小,即HTML元素的字体大小. 有一个比较普遍的误解,认为em单位是相对于父元素的字体大小.事实上,根据W3C ...
- scala的应用--UDF:用户自定义函数
在window10下安装了hadoop,用ida创建maven项目. <properties> <spark.version>2.2.0</spark.version&g ...
- WPF XAML的读法
XAML 一直以为读作X-A-M-L 不过 一直都是念错了 正确念法:ZAMMEL 类似:ZeIMO [平音]
- 最简单的 kubernetes 高可用安装方式
sealos 项目地址:https://github.com/fanux/sealos 本文教你如何用一条命令构建 k8s 高可用集群且不依赖 haproxy 和 keepalived,也无需 ans ...
- Java学习:网络编程总结
Java网络编程总结 一.概述 计算机网络是通过传输介质.通信设施和网络通信协议,把分散在不同地点的计算机设备互连起来,实现资源共享和数据传输的系统.网络编程就就是编写程序使联网的两个(或多个)设备( ...
- Centos7/Ubuntu 初始化硬盘分区、挂载
刚刚在腾讯云买了一台服务器,刚买的服务器的数据盘都是需要自己来分区的,下面就记录一下操作. 通过命令fdisk-l查看硬盘信息 可以看到有两块硬盘/dev/vda和/dev/vdb,启动vda是系统盘 ...
- Newtonsoft.Json使用技巧
本篇将为大家介绍Newtonsoft.Json的一些高级用法,可以修改很少的代码解决上述问题. 阅读目录 Newtonsoft.Json介绍 基本用法 高级用法 总结 回到顶部 Newtonsoft. ...