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. ORACLE中%TYPE和%ROWTYPE的使用

     1 %TYPE说明 为了使一个变量的数据类型与还有一个已经定义了的变量(尤其是表的某一列)的数据类型相一致,Oracle提供了%TYPE定义方式.当被參照的那个变量的数据类型改变了之后,这个新定 ...

  2. Java I/O— 梳理各种“流”

    背景 Java核心库java.io它提供了一个综合IO接口.包含:文件读写.标准装备输出等..Java在IO它是流为基础进行输入输出的.全部数据被串行化写入输出流,或者从输入流读入. -- 百度百科 ...

  3. 重温委托(delegate)和事件(event)

    1.delegate是什么 某种意义上来讲,你可以把delegate理解成C语言中的函数指针,它允许你传递一个类A的方法m给另一个类B的对象,使得类B的对象能够调用这个方法m,说白了就是可以把方法当作 ...

  4. Linux scp文件复制

    scp是 secure copy的缩写, scp是linux系统下基于ssh登陆进行安全的远程文件拷贝命令.  1.scp命令的用处:  scp在网络上不同的主机之间复制文件,它使用ssh安全协议传输 ...

  5. Inno Setup打包添加和去除管理员权限

    原文:Inno Setup打包添加和去除管理员权限 添加管理员权限 1.在[Setup]节点添加 PrivilegesRequired=admin 2.进入安装目录,找到文件SetupLdr.e32, ...

  6. Redis集群明细文档(转)

    相信很多用过Redis的同学都知道,Redis目前版本是没有提供集群功能的,只能单打独斗.如果要实现多台Redis同时提供服务只能通过客户端自身去实现.目前根据文档已经看到Redis正在开发集群功能, ...

  7. java读取XML文件的四种方式

    java读取XML文件的四种方式 Xml代码 <?xml version="1.0" encoding="GB2312"?> <RESULT& ...

  8. 开源一个适用iOS的数据库表结构更新机制的代码

    将前段时间开源的代码.公布一下: ARDBConfig On the iOS, provide a database table structure update mechanism, ensure ...

  9. NYOJ 104 最大子矩阵(二维DP)

    最大和 时间限制:1000 ms  |  内存限制:65535 KB 难度:5 描写叙述 给定一个由整数组成二维矩阵(r*c),如今须要找出它的一个子矩阵,使得这个子矩阵内的全部元素之和最大,并把这个 ...

  10. java获取当前日期时间代码总结

    1.获取当前时间,和某个时间进行比较.此时主要拿long型的时间值.  方法如下: 要使用 java.util.Date .获取当前时间的代码如下  代码如下 复制代码 Date date = new ...