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 

动规 状压DP

看到数据范围就是状压DP了吧233

有那么一阵子我有DFS可以剪枝强行卡过去的错觉,然而果然是错觉。

f[i]记录的是当前状态(状压当前有哪些牛)的最大安全因子是多大

 #include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int mxn=;
int read(){
int x=,f=;char ch=getchar();
while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>='' && ch<=''){x=x*-''+ch;ch=getchar();}
return x*f;
}
int f[<<];
int g[<<];
struct node{
int h,w,s;
}a[mxn];
int n,H;
int ans=-;
int main(){
int i,j;
n=read();H=read();int smm=;
for(i=;i<n;i++){
a[i].h=read();a[i].w=read();a[i].s=read();
smm+=a[i].h;
}
if(smm<H){printf("Mark is too tall\n");return ;}
memset(f,-,sizeof f);
f[]=INF;
int ed=(<<n)-;
for(i=;i<=ed;i++){
for(j=;j<n;j++){
if((i>>j)&)continue;
int v=i^(<<j);
if(f[i]<a[j].w)continue;
int t=min(f[i]-a[j].w,a[j].s);
f[v]=max(f[v],t);
g[v]=g[i]+a[j].h;//累计高度
if(v && g[v]>=H)ans=max(ans,f[v]);
}
}
if(ans==-)printf("Mark is too tall\n");
else printf("%d\n",ans);
return ;
}

洛谷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 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> ...

  3. LUOGU 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' ...

  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. 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& ...

  8. 洛谷P3110 [USACO14DEC]驮运Piggy Back

    P3110 [USACO14DEC]驮运Piggy Back 题目描述 贝西和她的妹妹艾尔斯白天在不同的地方吃草,而在晚上他们都想回到谷仓休息.聪明的牛仔,他们想出了一个计划,以尽量减少他们在步行时花 ...

  9. 洛谷 P3111 [USACO14DEC]牛慢跑Cow Jog_Sliver

    P3111 [USACO14DEC]牛慢跑Cow Jog_Sliver 题目描述 The cows are out exercising their hooves again! There are N ...

随机推荐

  1. TI C6000 数据存储处理与性能优化

    存储器之于CPU好比仓库之于车间.车间加工过程中的原材料.半成品.成品等均需入出仓库,生产效率再快,如果仓库周转不善,也必然造成生产阻塞.如同仓库需要合理地规划管理一般,数据存储也需要恰当的处理技巧来 ...

  2. 笔记-select,poll,epoll

    笔记-select,poll,epoll 1.      I/O多路复用 I/O多路复用是指:通过一种机制或一个进程,可以监视多个文件描述符,一旦描述符就绪(写或读),能够通知程序进行相应的读写操作. ...

  3. java跨服务器请求url获得数据

    在项目中,有时需要通过请求远程服务器上的url获取数据(前提是程序所在服务器可以和url服务器ping成功), 用java在后台发送请求时,用到了java.net.URL, java.net.URLC ...

  4. Linux命令学习总结(一)

    命令 -选项 参数 如果选项是一个单词时,选项前面要加2个- modprobe -r pcspkr   在终端中输入的时候有声音,可以用这个命令屏蔽声音 ,需要root权限 useradd userd ...

  5. 斐波那契数列(Fibonacci) iOS

    斐波那契数列Fibonacci 斐波那契数列指的是这样一个数列 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,377,610,987,1597,2 ...

  6. Pythontutor:可视化代码在内存的执行过程

    http://www.pythontutor.com/visualize.html今天去问开发一个Python浅拷贝的问题,开发给了一个神器,可以可视化代码在内存的执行过程,一看即懂,太NB了!~真是 ...

  7. adb启动和关闭

    启动adb服务: cmd("adb start-server"); 关闭adb服务: cmd("adb start-server");

  8. 第三方库的安装:Pangolin

    Pangolin: 一款开源的OPENGL显示库,可以用来视频显示.而且开发容易. 代码我们可以从Github 进行下载:https://github.com/stevenlovegrove/Pang ...

  9. Hadoop平台K-Means聚类算法分布式实现+MapReduce通俗讲解

        Hadoop平台K-Means聚类算法分布式实现+MapReduce通俗讲解 在Hadoop分布式环境下实现K-Means聚类算法的伪代码如下: 输入:参数0--存储样本数据的文本文件inpu ...

  10. React03 移动端跨平台开发

    目录 React-day03 RN移动端开发 了解React-Native 了解React-Native工作流程 创建第一个React-Native项目 * 了解React-Native项目及结构 开 ...