POJ 1276  Cash Machine（多重背包）

Cash Machine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24132		Accepted: 8446

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

题意， 给你最大的钱数 Case ，和 nk 种不同的面值, 每种面值D1...DN 对应的数量 n1....nN, 找出不超过最大钱数Case

思路：多重背包

import java.io.*;
import java.util.*;
/*
 *
 * author : deng_hui_long 
 * Date   : 2013-8-31
 *
*/
public class Main {
	int Case,n;
	int dp[]=new int[1000000];
	public static void main(String[] args) {
		new Main().work();
	}
	void work(){
		Scanner sc=new Scanner(new BufferedInputStream(System.in));
		while(sc.hasNext()){
			Case=sc.nextInt();
			n=sc.nextInt();
			if(n==0&&Case==0)
				break;
			Node node[]=new Node[n];
			Arrays.fill(dp,0);
			for(int i=0;i<n;i++){
				int a=sc.nextInt();
				int b=sc.nextInt();
				node[i]=new Node(a,b);
			}
			for(int i=0;i<n;i++){
				multiplePack(node[i].val,node[i].val,node[i].cost);
			}
			System.out.println(dp[Case]);
		}
	}
	//多重背包
	void multiplePack(int cost,int weight,int amount){
		if(cost*amount>=Case)//超过最大的钱数，按完全背包处理
			completePack(cost,weight);
		else{//小于最大的钱数，按01背包处理
			int k=1;
			while(k<amount){
				zeroOnePack(k*cost,k*weight);
				amount-=k;
				k<<=1;//左移一位，表示乘以2
			}
			zeroOnePack(amount*cost,amount*weight);
		}
	}
	//完全背包
	void completePack(int cost,int weight){
		for(int i=cost;i<=Case;i++){
			dp[i]=Math.max(dp[i],dp[i-cost]+weight);
		}
	}
	//01背包
	void zeroOnePack(int cost,int weight){
		for(int i=Case;i>=cost;i--)
			dp[i]=Math.max(dp[i],dp[i-cost]+weight);
	}
	class Node{
		int cost;
		int val;
		Node(int cost,int val){
			this.cost=cost;
			this.val=val;
		}
	}
}