POJ 1276 Cash Machine 多重背包--二进制优化

浏览数：20 / 时间：2015年06月20日

Cash Machine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 28337		Accepted: 10113

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

有n种不同面值的货币，告诉你每种货币的面值和数量，找出利用这些货币可以凑成的最接近且小于等于给定的数字cash的金额。

多重背包问题。

可以转换成01背包，再进行二进制优化，时间复杂度就由O(V*Σn[i])降为O(V*Σlog n[i])。详见背包九讲。

//600K	79MS
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[100007];
int weight[10007],value[10007];
int main()
{
    int sum,n;
    while(scanf("%d%d",&sum,&n)!=EOF)
    {
        int a,b,num=0;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)//进行二分优化
        {
            scanf("%d%d",&a,&b);
            for(int j=1;j<=a;j<<=1)
            {
                value[num]=weight[num]=j*b;
                num++;
                a-=j;
            }
            if(a>0)
            {
                value[num]=weight[num]=a*b;
                num++;
            }
        }
        for(int i=0;i<num;i++)
            for(int j=sum;j>=weight[i];j--)
                dp[j]=max(dp[j-weight[i]]+value[i],dp[j]);
         printf("%d\n",dp[sum]);
    }
    return 0;
}

背包九讲中的二进制优化：

//524K	63MS
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[100007],num[17],weight[17];
int sum;
void ZeroOnePack(int cost)
{
    for(int i=sum;i>=cost;i--)
        dp[i]=max(dp[i],dp[i-cost]+cost);
}
void CompletePack(int cost)
{
    for(int i=cost;i<=sum;i++)
        dp[i]=max(dp[i],dp[i-cost]+cost);
}
void MultiplePack(int count,int cost)
{
    if(count*cost>sum)CompletePack(cost);
    else
    {
        int k=1;
        while(k<count)
        {
            ZeroOnePack(k*cost);
            count-=k;
            k<<=1;
        }
        ZeroOnePack(count*cost);
    }
}
int main()
{
    int n;
    while(scanf("%d%d",&sum,&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d%d",&num[i],&weight[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            MultiplePack(num[i],weight[i]);
        printf("%d\n",dp[sum]);
    }
}

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POJ 1276 Cash Machine 多重背包--二进制优化