背包（2）poj1276 Cash Machine（多重背包）

Cash Machine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22943		Accepted: 8038

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

735
630
0
0

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

 

 

/*
题意：给你几种面额的钱币和数量，和一个标准cash，问你用所给的钱币，能够得到小于等于cash值的最大值。	明显的多重背包。
方法：将多重背包化为01背包，可以另开一个数组存每种钱币的个数，在做01背包时多加一个条件就行了。
	还可以用二进制方法，把数目用二进制进行分解，比如5个物品，分为1个，2个和2个，用这三个新的物品可以通过01背包的方式重新组合成1,2,3,4,5个物品的情况；
	7个物品可以分为1个，2个和4个……
*/
/* 法一：记录每种钱币的个数 1472K 16MS */
#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
using namespace std;
int val[11],num[11];
int ans[100005];
int h[100005];
int main()
{
	int n,cash,a;
	while(scanf("%d%d",&cash,&n) != EOF)
	{
		for(int i=1; i<=n; i++)
			scanf("%d%d", &num[i], &val[i]);
		if(cash == 0 || n == 0)
		a = 0;
		else{
		memset(ans, 0, sizeof(ans));
		for(int i=1; i<=n; i++)
		{
			memset(h, 0, sizeof(h));
			for(int j=1; j<=cash; j++)
			{
				if(j - val[i] >= 0 && h[j - val[i]] < num[i] && ans[j] < ans[j-val[i]] + val[i])
				{
					ans[j] = ans[j-val[i]] + val[i];
					h[j] = h[j-val[i]] + 1;
				}
			}
		}
		a = ans[cash];
		}
		printf("%d\n",a);
	}
	return 0;
}
/* 法二：二进制处理 776K 63MS */
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int c[200];
int ans[100005];
int main()
{
	int cash,N,n,d,k;
	while(~scanf("%d%d",&cash,&N))
	{
		memset(c,0,sizeof(c));
		k = 0;
		for (int i = 0; i < N; i++)
		{
			scanf("%d%d",&n,&d);
			if(n==0 || d==0)	continue;
			int x = 1,tem = n,sum = 0;
			while(n>1)
			{
				c[k++] = x*d;
				sum += x;
				n/=2;	x*=2;
			}
			if(tem>sum)
			c[k++] = (tem-sum)*d;			
		}
		memset(ans,0,sizeof(ans));
		for (int i = 0; i < k; i++)
		{
 			for (int j = cash; j >=c[i] ; j--)
			{
				ans[j] = max(ans[j],ans[j-c[i]]+c[i]);
			}
		}
		printf("%d\n",ans[cash]);
	}
	return 0;
}