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. Docker系列(十五):Openshift 简介

    1.简单了解openshift相关组件 1.openshift是基于容器技术构建的一个云平台 2.kubernetes是容器编排组件 3.docker是容器引擎驱动组件 4.openshift在Pas ...

  2. mybatis第二篇—参数绑定

    不管我们在做数据库作业或者任务还是当时的仅靠jdbc来写一个管理系统的时候,sql语句需要一些参数,从而来实现模糊查询,精确查询,插入数据,更新数据和删除数据.这些参数,在mybatis里面,又该如何 ...

  3. ES6之数值的扩展学习

    引自:http://es6.ruanyifeng.com/#docs/number 二进制和八进制表示法 ES6 提供了二进制和八进制数值的新的写法,分别用前缀0b(或0B)和0o(或0O)表示. 0 ...

  4. HTML5定位功能,实现在百度地图上定位

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  5. Errors were encountered while processing: mysql-server-5.5

    ubuntu 中运行完sudo apt-get install curl之后,最后出现: ldconfig deferred processing now taking place Errors we ...

  6. Ionic Cordova Sqlite 实现保存用户名登陆

    1.添加sqlite 组件 cordova plugin add https://github.com/litehelpers/Cordova-sqlite-storage.git --save 2. ...

  7. React 组件&Props

    组件&Props 组件&Props 组件可以将UI切分成一些独立的.可复用的部件,这样你就只需要专注于构建每一个单独的组件. 组件从概念上看就像是函数,它可以接受任意的输入值(称之为& ...

  8. PHP实现记录浏览历史页面

    <?php /******* 说明:cookie只能保存字符串 本实例中,需要保存多个URL(历史访问记录),思路是先将URL数组转为字符串,然后保存,读取时,再循环读取 *******/ // ...

  9. AppbarLayout的简单用法

    在许多App中看到, toolbar有收缩和扩展的效果, 例如:   appbar.gif 要实现这样的效果, 需要用到: CoordinatorLayout和AppbarLayout的配合, 以及实 ...

  10. 【BZOJ4161】Shlw loves matrixI

    题目描述 给定数列 {hn}前k项,其后每一项满足 hn = a1h(n-1) + a2h(n-2) + ... + ak*h(n-k) 其中 a1,a2...ak 为给定数列.请计算 h(n),并将 ...