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. HTML+CSS : H5+CSS3

    HTML5语义化标签: header nav(导航) article section(章节) aside(侧边栏) footer------------------------------------ ...

  2. B1081 检查密码 (15分)

    B1081 检查密码 (15分) 本题要求你帮助某网站的用户注册模块写一个密码合法性检查的小功能.该网站要求用户设置的密码必须由不少于6个字符组成,并且只能有英文字母.数字和小数点 .,还必须既有字母 ...

  3. python基础之闭包函数和装饰器

    补充:全局变量声明及局部变量引用 python引用变量的顺序: 当前作用域局部变量->外层作用域变量->当前模块中的全局变量->python内置变量 global关键字用来在函数或其 ...

  4. Hive 数据实战

    需求 remote_addr 用户IP 1.用于根据地址确认区域 2.用于统计来自同一个(外网)用户的访问数量 time_local 用户访问时间 1.分析用户访问时间段 2.合理安排客服上班时间 r ...

  5. linux下解压命令大全(转)

    .tar 解包:tar xvf FileName.tar打包:tar cvf FileName.tar DirName(注:tar是打包,不是压缩!)———————————————.gz解压1:gun ...

  6. 7,MongoDB 之 Limit 选取 Skip 跳过 Sort 排序

    我们已经学过MongoDB的 find() 查询功能了,在关系型数据库中的选取(limit),排序(sort) MongoDB中同样有,而且使用起来更是简单 首先我们看下添加几条Document进来 ...

  7. python 发送 get post请求

    GET请求: python2.7: import urllib,urllib2 url='http://192.168.199.1:8000/mainsugar/loginGET/' textmod ...

  8. spring里面的context:component-scan

    原文:http://jinnianshilongnian.iteye.com/blog/1762632 component-scan的作用的自动扫描,把扫描到加了注解Java文件都注册成bean &l ...

  9. Eclipse 创建 Java 接口---Eclipse教程第11课

    打开新建 Java 接口向导 新建 Java 接口向导可以创建新的 Java 接口.打开向导的方式有: 点击 File 菜单并选择 New > Interface 在 Package Explo ...

  10. linux shell 总结 (整理)

    ls /usr/bin/ info #路径操作 dirname basename #“”和‘’与 ` ` 在shell变量中的区别 “ ” 允许通过$符引用其他变量 ‘’禁止引用其他变量符,视为普通字 ...