题目描述

Farmer John and his herd are playing frisbee. Bessie throws the

frisbee down the field, but it’s going straight to Mark the field hand

on the other team! Mark has height H (1 <= H <= 1,000,000,000), but

there are N cows on Bessie’s team gathered around Mark (2 <= N <= 20).

They can only catch the frisbee if they can stack up to be at least as

high as Mark. Each of the N cows has a height, weight, and strength.

A cow’s strength indicates the maximum amount of total weight of the

cows that can be stacked above her.

Given these constraints, Bessie wants to know if it is possible for

her team to build a tall enough stack to catch the frisbee, and if so,

what is the maximum safety factor of such a stack. The safety factor

of a stack is the amount of weight that can be added to the top of the

stack without exceeding any cow’s strength.

FJ将飞盘抛向身高为H(1 <= H <= 1,000,000,000)的Mark,但是Mark

被N(2 <= N <= 20)头牛包围。牛们可以叠成一个牛塔,如果叠好后的高度大于或者等于Mark的高度,那牛们将抢到飞盘。

每头牛都一个身高,体重和耐力值三个指标。耐力指的是一头牛最大能承受的叠在他上方的牛的重量和。请计算牛们是否能够抢到飞盘。若是可以,请计算牛塔的最大稳定强度,稳定强度是指,在每头牛的耐力都可以承受的前提下,还能够在牛塔最上方添加的最大重量。

输入输出格式

输入格式:

INPUT: (file guard.in)

The first line of input contains N and H.

The next N lines of input each describe a cow, giving its height,

weight, and strength. All are positive integers at most 1 billion.

输出格式:

OUTPUT: (file guard.out)

If Bessie’s team can build a stack tall enough to catch the frisbee, please output the maximum achievable safety factor for such a stack.

Otherwise output “Mark is too tall” (without the quotes).

输入输出样例

输入样例#1:

4 10

9 4 1

3 3 5

5 5 10

4 4 5

输出样例#1:

2

解题思路

乍一看这道题是个以前讲过的贪心,就是按照力量和重量排序,但是这个还有个高度,并且问的是最大稳定,所以那种方法似乎不行。考虑状压,dp[S]表示选的状态为S时的最大承重,gg[S]表示所选为S时的最大高度,可以提前预处理出来。转移方程dp[S]=max(dp[S],min(dp[S^(1<

代码

#include<iostream>
#include<cstdio>
#include<cstring> using namespace std;
const int MAXN = 21;
typedef long long LL; int H,n;
int dp[1<<MAXN],gg[1<<MAXN];
int w[MAXN],h[MAXN],a[MAXN];
int ans=-1; int main(){
memset(dp,-0x3f,sizeof(dp));dp[0]=0x3f3f3f3f;
scanf("%d%d",&n,&H);
for(register int i=1;i<=n;i++)
scanf("%d%d%d",&h[i],&w[i],&a[i]);
for(register int S=0;S<1<<n;S++)
for(register int i=1;i<=n;i++)
if(S&(1<<i-1)) gg[S]+=h[i];
for(register int S=0;S<1<<n;S++){
for(register int i=1;i<=n;i++)if(((S&(1<<i-1))))
dp[S]=max(dp[S],min(a[i],dp[S^(1<<i-1)]-w[i]));
if(gg[S]>=H && dp[S]>=0) ans=max(ans,dp[S]);
}
if(ans==-1) puts("Mark is too tall");
else printf("%d",ans);
return 0;
}

LUOGU P3112 [USACO14DEC]后卫马克Guard Mark的更多相关文章

  1. 洛谷 P3112 [USACO14DEC]后卫马克Guard Mark

    题目描述 Farmer John and his herd are playing frisbee. Bessie throws the frisbee down the field, but it' ...

  2. 洛谷P3112 [USACO14DEC]后卫马克Guard Mark

    题目描述 Farmer John and his herd are playing frisbee. Bessie throws the frisbee down the field, but it' ...

  3. 洛谷 3112 [USACO14DEC]后卫马克Guard Mark——状压dp

    题目:https://www.luogu.org/problemnew/show/P3112 状压dp.发现只需要记录当前状态的牛中剩余承重最小的值. #include<iostream> ...

  4. [Luogu3112] [USACO14DEC]后卫马克Guard Mark

    题意翻译 FJ将飞盘抛向身高为H(1 <= H <= 1,000,000,000)的Mark,但是Mark被N(2 <= N <= 20)头牛包围.牛们可以叠成一个牛塔,如果叠 ...

  5. [USACO14DEC]后卫马克Guard Mark

    题目描述 FJ将飞盘抛向身高为H(1 <= H <= 1,000,000,000)的Mark,但是Mark 被N(2 <= N <= 20)头牛包围.牛们可以叠成一个牛塔,如果 ...

  6. 洛谷 P3112 后卫马克Guard Mark

    ->题目链接 题解: 贪心+模拟 #include<algorithm> #include<iostream> #include<cstring> #incl ...

  7. 【题解】Luogu P3110 [USACO14DEC]驮运Piggy Back

    [题解]Luogu P3110 [USACO14DEC]驮运Piggy Back 题目描述 Bessie and her sister Elsie graze in different fields ...

  8. 洛谷 P3112 后卫马克 —— 状压DP

    题目:https://www.luogu.org/problemnew/show/P3112 状压DP...转移不错. 代码如下: #include<iostream> #include& ...

  9. bzoj 3824: [Usaco2014 Dec]Guard Mark【状压dp】

    设f[s]为已经从上到下叠了状态为s的牛的最大稳定度,转移的话枚举没有在集合里并且强壮度>=当前集合牛重量和的用min(f[s],当前放进去的牛还能承受多种)来更新,高度的话直接看是否有合法集合 ...

随机推荐

  1. PHP面向对象魔术方法之__get 和 __set函数

    l 基本的介绍 (1) 当我们去使用不可以访问的属性时,系统就会调用__get方法. (2) 不可以访问的属性指的是(1 . 该属性不存在 2. 直接访问了protected或者private属性) ...

  2. My solutions to the exercises in "The Boost C++ Libraries"

    I like books with excercises, but I also want solutions to see if I got it right. When working throu ...

  3. C#控件的闪烁问题解决方法总结

    最近对代码作了一些优化,试验后效果还可以,但是发现界面会闪烁,具体是TreeView控件会闪烁,语言为C#,IDE为VS2005.在查阅一些资料,使用了一些基本技术后(如开启双缓冲),发现没什么效果. ...

  4. 使用scrapy框架来进行抓取的原因

    在python爬虫中:使用requests + selenium就可以解决将近90%的爬虫需求,那么scrapy就是解决剩下10%的吗? 这个显然不是这样的,scrapy框架是为了让我们的爬虫更强大. ...

  5. tensorflow+inceptionv3图像分类网络结构的解析与代码实现

    tensorflow+inceptionv3图像分类网络结构的解析与代码实现 论文链接:论文地址 ResNet传送门:Resnet-cifar10 DenseNet传送门:DenseNet SegNe ...

  6. 乐观、悲观锁、redis分布式锁

    悲观锁总是假设最坏的情况,每次去拿数据的时候都认为别人会修改,所以每次在拿数据的时候都会上锁,这样别人想拿这个数据就会阻塞直到它拿到锁(共享资源每次只给一个线程使用,其它线程阻塞,用完后再把资源转让给 ...

  7. Yaf--个人封装yaf的框架+swoole+elasticsearch(Window+linux版)

    这是基于c写底层的yaf框架集成PDO+predis+读写分离+composer+全局异常处理+多模块开发+Log日志记录简单容易上手的框架 注意:window版没有swoole和Smarty主要用作 ...

  8. 【JAVA】Class.getResource()与ClassLoader.getResource()的区别

    转载自:https://blog.csdn.net/qq_33591903/article/details/91444342 Class.getResource()与ClassLoader.getRe ...

  9. Lamdba表达式的代码使用讲解

    public class Lambda{ public static void main(String[] args) { repeat(10, (i)->System.out.print(&q ...

  10. UOJ#80. 二分图最大权匹配 模板

    #80. 二分图最大权匹配 描述 提交 自定义测试 从前一个和谐的班级,有 nlnl 个是男生,有 nrnr 个是女生.编号分别为 1,…,nl1,…,nl 和 1,…,nr1,…,nr. 有若干个这 ...