Cash Machine 取款机多重背包优化成01背包

最新推荐文章于 2019-07-28 18:48:56 发布

原创最新推荐文章于 2019-07-28 18:48:56 发布 · 174 阅读

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本文深入探讨了一种典型的多重背包问题，即如何计算银行提款机在有限的钞票面额和数量下，能提供的最大现金金额，以满足用户的取款需求。文章通过一个具体的示例，详细介绍了使用二进制优化和01背包算法来解决这一问题的方法。

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题意：

A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example,

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.

输入：

The program input is from standard input. Each data set in the input stands for a particular transaction and has the format:

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.

输出：

For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.

样例输入：

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

样例输出：

提示：

The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount of requested cash.

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.

做的背包里还是太少，这样一个模板题比赛的时候都没做出来，路还还长，多加努力把。

大体题意就是给出一个用户所需要取出的货币数，给出银行可以提供的货币的面额，每种货币都有有限的数量，输出银行所能提供的最大的货币。纯粹的多重背包模板题。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100000+5;
int a[maxn],dp[maxn];
int main()
{

    int n,m;
    while(cin>>n>>m)
    {
        int cnt=0;
        memset(dp,0,sizeof(dp));
       for(int i=0;i<m;i++)
       {
           int x,y;
           cin>>x>>y;
           for(int j=1;;j*=2)//二进制优化转化为01背包
           {
               if(x>j)
               {
                   a[cnt++]=j*y;
                   x-=j;
               }
               else
               {
                   a[cnt++]=x*y;
                   break;
               }
           }
       }
       for(int i=0;i<cnt;i++)//01背包模板
        for(int j=n;j>=a[i];j--)
       {
           dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
       }
        cout<<dp[n]<<endl;
    }
    return 0；
}