Cash Machine----POJ_1276----多重背包

最新推荐文章于 2019-12-01 17:08:05 发布

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本文解析了一道经典的CashMachine问题，该问题涉及到多重背包算法的应用。在一个具有多种不同面值货币的银行中，每种面值有若干张。目标是在不超过给定数额的情况下，求出能够提供的最大金额。文章通过示例详细介绍了如何使用多重背包算法解决这个问题。

题目地址：http://poj.org/problem?id=1276

Cash Machine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19785		Accepted: 6892

Description

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.

Input

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.

Output

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

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

Hint

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.

Source

Southeastern Europe 2002

本题的题意就是，一个银行里面有若干种不同面值的货币，每种有若干张。现在有一个最大数字给你，然后要你求出接近于这个数字，但是要小于等于它的最大值。这个毫无疑问是一个多重背包问题。背包的容量就是给的这个最大数字，然后要求出背包所能装下的最大价值，即所求。多重背包不懂的可参考： http://love-oriented.com/pack/P03.html

#include<iostream>
using namespace std;

#define MAX 100010
#define MAX2 20
int f[MAX];
int V;


void CompletePack(int c,int w)
{
	int v;
	for(v=c;v<=V;v++)
	{
		if(f[v-c]+w > f[v])
			f[v] = f[v-c]+w;
	}
}

void ZeroOnePack(int c,int w)
{
	int v;
	for(v=V;v>=c;v--)
	{
		if(f[v-c]+w > f[v])
			f[v] = f[v-c]+w;
	}
}

void MulPack(int c,int w,int m)
{
	if(c*m >= V)
	{
		CompletePack(c,w);
		return ;
	}

	int k=1;

	while(k <= m)
	{
		ZeroOnePack(k*c,k*w);
		m = m-k;
		k=k*2;
	}

	ZeroOnePack(m*c,m*w);
}

int main()
{
	int c[MAX2];
    int w[MAX2];
	int m[MAX2];
	while(scanf("%d",&V) != EOF)
	{
		int n;
		scanf("%d",&n);
		int i;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&m[i],&w[i]);
			c[i] = w[i];
		}
		if(V == 0)
		{
			printf("0\n");
		}
		else
		{
			for(i=0;i<=V;i++)
				f[i] = 0;
			for(i=1;i<=n;i++)
				MulPack(c[i],w[i],m[i]);
			printf("%d\n",f[V]);
		}
	}


	return 0;
}