HDOJ-三部曲-多重背包-1014-Cash Machine
通过这道题我基本了解了利用二进制对多重背包问题进行优化的思想。
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24191 | Accepted: 8466 |
Description
N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10
means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each.
Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.
Notes: @ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.
Input
cash N n1 D1 n2 D2 ... nN DN
where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers in the input. The input data are correct.
Output
Sample Input
735 3 4 125 6 5 3 350
633 4 500 30 6 100 1 5 0 1
735 0
0 3 10 100 10 50 10 10
Sample Output
735
630
0
0
Hint
In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered cash.
In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash.
Source
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int cash,n;
while(cin>>cash>>n)
{
int i,j,k=0,num[11],den[11],sc[10000];
int *dp=new int[100001];
memset(dp,0,(cash+1)*sizeof(int));
for(i=1;i<=n;i++)
{
cin>>num[i]>>den[i];
int t=1,amount=num[i];
while(amount>0) //用二进制思想进行优化,化为01背包问题
{
if(amount>=t)
{
sc[k++]=t*den[i];
amount-=t;
t*=2;
}
else
t=1;
}
}
/* for(i=0;i<k;i++)
cout<<sc[i]<<' ';
cout<<endl;*/
for(i=0 ;i<k;i++)
for(j=cash;j>=sc[i];j--)
dp[j]=max(dp[j],dp[j-sc[i]]+sc[i]);
cout<<dp[cash]<<endl;
delete dp;
}
}
HDOJ-三部曲-多重背包-1014-Cash Machine的更多相关文章
- hdoj 2191(多重背包)
悼念512汶川大地震遇难同胞——珍惜现在,感恩生活 Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/ ...
- hdu 2191 (多重背包+二进制优化)
Problem Description 急!灾区的食物依然短缺!为了挽救灾区同胞的生命,心系灾区同胞的你准备自己采购一些粮食支援灾区,现在假设你一共有资金n元,而市场有m种大米,每种大米都是袋装产品, ...
- Poj 1276 Cash Machine 多重背包
Cash Machine Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 26172 Accepted: 9238 Des ...
- poj 1276 Cash Machine(多重背包)
Cash Machine Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 33444 Accepted: 12106 De ...
- POJ1276:Cash Machine(多重背包)
Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver ap ...
- Cash Machine(多重背包二进制转换)
个人心得:多重背包,自己根据转换方程写总是TLE,后面去网上看了二进制转换,不太理解: 后面仔细想了下,用自己的思想理解下把,就是将对应number,cash总和用二进制拆分, 然后全部装入到一个数组 ...
- POJ 1276 Cash Machine(单调队列优化多重背包)
Cash Machine Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 38986 Accepted: 14186 De ...
- POJ 1276:Cash Machine 多重背包
Cash Machine Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 30006 Accepted: 10811 De ...
- POJ1276 - Cash Machine(多重背包)
题目大意 给定一个容量为M的背包以及n种物品,每种物品有一个体积和数量,要求你用这些物品尽量的装满背包 题解 就是多重背包~~~~用二进制优化了一下,就是把每种物品的数量cnt拆成由几个数组成,1,2 ...
- 【转载】poj 1276 Cash Machine 【凑钱数的问题】【枚举思路 或者 多重背包解决】
转载地址:http://m.blog.csdn.net/blog/u010489766/9229011 题目链接:http://poj.org/problem?id=1276 题意:机器里面共有n种面 ...
随机推荐
- hdu----(2222)Keywords Search(ac自动机)
Keywords Search Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)T ...
- 模拟器的tableView的分割线不显示
只有iOS9和iPhone6 plus模拟器上TableView分割线不会显示. 原因: 由于iPhone6 plus的分辨率较高,开发的时候同常都使用command + 3 或者 command + ...
- Rudolph javascript 监听简单对象属性的变化 -- 回调函数的应用
http://www.oschina.net/code/snippet_1590754_46481 //简单对象的属性的变化监控 //通过setAttr改变属性的值 var o = { 'a':2, ...
- Java script基础
Java script基础 Js的每个语句后面都要有分号. <script type="text/java script">所有JS内容</script> ...
- while、do while练习——7月24日
while循环的格式是for循环的变形 //while 循环(当循环),是for循环的变形 //for(int i=0;i<=5;i++) //{ // Console.WriteLine(&q ...
- Objective-C:Foundation框架-概述
iOS的整体架构(以iOS8为例)图如下: 从Cocoa Touch到Core OS下面四层包含了开发iOS应用程序所用到的所有API(第三方框架也是基于这几个层的).每个层又都包含了许多框架.框架就 ...
- JVM调优(这里主要是针对优化基于分布式Mahout的推荐引擎)
优化推荐系统的JVM关键参数 -Xmx 设定Java允许使用的最大堆空间.例如-Xmx512m表示堆空间上限为512MB -server 现代JVM有两个重要标志:-client和-server,分别 ...
- [Js]无缝滚动
效果: 1.默认缓慢往左滚动 2.放到左箭头上还是向左滚动,放到右箭头上向右滚动 3.放到图片上停止滚动,移出继续滚动 思路: 1.计算图片列表ul的宽度 2.开启定时器,使其向左边距不断增大,造成向 ...
- ssh curl 命令理解
使用一条命令抓取一本小说 curl "http://www.23hh.com/book/1/1019/"|iconv -c -f gbk -t utf8 |sed 's/" ...
- 数据结构-Stack和Queue
实现: #include "c2_list.h" template <typename object> class Stack{ public: bool isEmpt ...