zhx and contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 455    Accepted Submission(s): 158

Problem Description
As one of the most powerful brushes in the world, zhx usually takes part in all kinds of contests.

One day, zhx takes part in an contest. He found the contest very easy for him.

There are n problems
in the contest. He knows that he can solve the ith problem
in ti units
of time and he can get vi points.

As he is too powerful, the administrator is watching him. If he finishes the ith problem
before time li,
he will be considered to cheat.

zhx doesn't really want to solve all these boring problems. He only wants to get no less than w points.
You are supposed to tell him the minimal time he needs to spend while not being considered to cheat, or he is not able to get enough points. 

Note that zhx can solve only one problem at the same time. And if he starts, he would keep working on it until it is solved. And then he submits his code in no time.
 
Input
Multiply test cases(less than 50).
Seek EOF as
the end of the file.

For each test, there are two integers n and w separated
by a space. (1≤n≤30, 0≤w≤109)

Then come n lines which contain three integers ti,vi,li.
(1≤ti,li≤105,1≤vi≤109)
 
Output
For each test case, output a single line indicating the answer. If zhx is able to get enough points, output the minimal time it takes. Otherwise, output a single line saying "zhx is naive!" (Do not output quotation marks).
 
Sample Input
1 3
1 4 7
3 6
4 1 8
6 8 10
1 5 2
2 7
10 4 1
10 2 3
 
Sample Output
7
8
zhx is naive!
 
Source
 

/* ***********************************************
Author :CKboss
Created Time :2015年03月19日 星期四 10时26分15秒
File Name :HDOJ5188.cpp
************************************************ */ #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map> using namespace std; const int maxn=200100; int n,w;
int sumtime,sumv;
int dp[maxn*30]; struct PB
{
int t,v,l;
}pb[50]; bool cmp(PB a,PB b)
{
return a.l-a.t<b.l-b.t;
} int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout); while(scanf("%d%d",&n,&w)!=EOF)
{
sumv=0;
for(int i=0;i<n;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
pb[i]=(PB){x,y,z};
sumv+=y;
}
if(sumv<w)
{
puts("zhx is naive!");
continue;
} sort(pb,pb+n,cmp);
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
for(int j=maxn;j>=max(pb[i].t,pb[i].l);j--)
dp[j]=max(dp[j],dp[j-pb[i].t]+pb[i].v); int ans=0;
for(int i=0;i<maxn;i++)
if(dp[i]>=w) { ans=i; break; } printf("%d\n",ans);
} return 0;
}

版权声明:来自: 代码代码猿猿AC路 http://blog.csdn.net/ck_boss

HDOJ 5188 zhx and contest 贪婪+01背包的更多相关文章

  1. HDOJ(HDU).3466 Dividing coins ( DP 01背包 无后效性的理解)

    HDOJ(HDU).3466 Dividing coins ( DP 01背包 无后效性的理解) 题意分析 要先排序,在做01背包,否则不满足无后效性,为什么呢? 等我理解了再补上. 代码总览 #in ...

  2. HDOJ(HDU).2546 饭卡(DP 01背包)

    HDOJ(HDU).2546 饭卡(DP 01背包) 题意分析 首先要对钱数小于5的时候特别处理,直接输出0.若钱数大于5,所有菜按价格排序,背包容量为钱数-5,对除去价格最贵的所有菜做01背包.因为 ...

  3. HDOJ(HDU).2602 Bone Collector (DP 01背包)

    HDOJ(HDU).2602 Bone Collector (DP 01背包) 题意分析 01背包的裸题 #include <iostream> #include <cstdio&g ...

  4. hdu 5188 zhx and contest [ 排序 + 背包 ]

    传送门 zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Othe ...

  5. ZOJ 3703 Happy Programming Contest(0-1背包)

    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3703 Happy Programming Contest Time Lim ...

  6. HDOJ 1203 I NEED A OFFER!(01背包)

    10397507 2014-03-25 23:30:21 Accepted 1203 0MS 480K 428 B C++ 泽泽 题目链接:http://acm.hdu.edu.cn/showprob ...

  7. Gym 101102A Coins -- 2016 ACM Amman Collegiate Programming Contest(01背包变形)

    A - Coins Time Limit:3000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Descript ...

  8. 饭卡------HDOJ杭电2546(还是01背包!!!!!!)

    Problem Description 电子科大本部食堂的饭卡有一种非常诡异的设计,即在购买之前推断剩余金额. 假设购买一个商品之前,卡上的剩余金额大于或等于5元,就一定能够购买成功(即使购买后卡上剩 ...

  9. HDU 5188 zhx and contest(带限制条件的 01背包)

    Problem Description As one of the most powerful brushes in the world, zhx usually takes part in all ...

随机推荐

  1. Android开发 更改返回button的图标

    非常多的Android应用左上角都有返回button 在默认的情况下 ADT会默认给一个返回图标 而作为开发需求 非常多都要求定制一个新的图标 在Android的站点上 发现了2种能够更改的方法 1. ...

  2. 跟我extjs5(38--单个模块的设计[6获得模块列表数据])

    跟我extjs5(38--单个模块的设计[6获得模块列表数据])         在程序的前一个表以及,据的执行过程. 在菜单中选择 "系统管理"--"模块分组" ...

  3. 【ArcGIS 10.2新特性】ArcGIS 10.2 for Server新特性

    ArcGIS forServer相关的很多重大消息,如与Portal for ArcGIS进行了集成,提供对实时数据支持.离线地图使用. 1.与Portal for ArcGIS集成 用户能够配置Po ...

  4. c#与oracle数据库连接池

    c#与oracle数据库连接池 在做一个项目,中间要使用webservice和oracle数据库.我在服务端做了用户身份认证,也就是使用session传递用户的登陆信息.在测试时,当用户少的时候,没有 ...

  5. Android数字签名解析(二)

    在Android数字签名解析(一)中,介绍了android进行签名的两种方式,当中用密钥对进行签名用到了signapk.jar这个java库. 以下我们就看看signapk签名实现过程,signapk ...

  6. 用bytecode来看try-catch-finally和return

    之前看过一篇关于return和finally运行顺序的文章.仅在Java的语言层面做了分析.事实上我倒认为直接看bytecode可能来的更清晰一点. 近期一直在看Java虚拟机规范.发现直接分析byt ...

  7. Android:主题(Theme)

    1.主题和样式的区别主要区别在 主题不能作用于单个View组建,主题应该对整个应用中的所有Activity起作用或者对指定的Activity起作用. 主题定义的格式应该是改变窗口的外观格式,例如窗口变 ...

  8. Windows phone 8 学习笔记(8) 定位地图导航

    原文:Windows phone 8 学习笔记(8) 定位地图导航 Windows phone 8 已经不使用自家的bing地图,新地图控件可以指定制图模式.视图等.bing地图的定位误差比较大,在模 ...

  9. String,StringBuffer以及StringBuilder的差别

    1.定义: String:String 类代表字符串. Java 程序中的全部字符串字面值(如"abc" )都作为此类的实例实现. 字符串是常量:它们的值在创建之后不能更改. 字符 ...

  10. New Hire Training Experience

    Game Description: 1. In a closed door, there will be a circle, and 30 numbers in the circle. 2. Each ...