经典算法题05-完全背包问题

完全背包(CompletePack)

问题

有N种物品和一个容量为V的背包，每种物品都有无限件可用。第i种物品的费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。

完全背包按其思路仍然可以用一个二维数组来写出：

f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i]|0<=k*c[i]<=v}

例题

问题来源：题目

Problem Description

Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

 

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it’s weight in grams.

 

Output

Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.".

 

Sample Input

    
    
    
    

     
     
     
     3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.
    
    
    
    

 

Source

Central Europe 1999

分析

同样可以转换成一维数组来表示:
伪代码如下:

for i=1..N
    for v=0..V
        f[v]=max{f[v],f[v-c[i]]+w[i]}

想必大家看出了和01背包的区别，这里的内循环是顺序的，而01背包是逆序的。
现在关键的是考虑：

为何完全背包可以这么写？

我们先来回忆下，01背包逆序的原因？是为了是max中的两项是前一状态值，这就对了。

那么这里，我们顺序写，这里的max中的两项当然就是当前状态的值了，为何？
因为每种背包都是无限的。

当我们把i从1到N循环时，f[v]表示容量为v在前i种背包时所得的价值，这里我们要添加的不是前一个背包，而是当前背包。所以我们要考虑的当然是当前状态。

答案

C/C++版：

#include<stdio.h>
int num[10001],w[500],v[500];
main()
{
    int n,m,e,f,t,i,j;
    for(scanf("%d",&t);t>0;t--)
    {
        scanf("%d%d",&e,&f);
        m=f-e;
        for(scanf("%d",&n),i=0;i<n;i++)
            scanf("%d%d",&v[i],&w[i]);
        num[0]=0;
        for(i=1;i<=m;i++)
            num[i]=-1;
        for(i=0;i<n;i++)
        {
            for(j=w[i];j<=m;j++)
            {
                if(num[j-w[i]]!=-1&&num[j]!=-1)
                {if(num[j-w[i]]+v[i]<num[j]) num[j]=num[j-w[i]]+v[i];}
                else if(num[j-w[i]]!=-1&&num[j]==-1)
                {num[j]=num[j-w[i]]+v[i];}
            }
        }
        if(num[m]!=-1)
        printf("The minimum amount of money in the piggy-bank is %d.\n",num[m]);
        else
            printf("This is impossible.\n");
    }
}

java版：

 public static void main(String[] args) throws IOException {

        int nCases=0;
        int nPack, nVolume1, nVolume2, nVolume;
        int weight[]= new int[510];
        int value[] = new int[510];
        int record[]= new int[1000];
        int INF = 1000000001;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String read = null;
        System.out.print("输入数据：");
        try {
            read = br.readLine();
            nCases = Integer.valueOf(read);
        } catch (IOException e) {
            e.printStackTrace();
        }
        System.out.println("输入的数据："+read);
        while(nCases-->0)
        {
            nVolume1 = Integer.valueOf(br.readLine());
            nVolume2 = Integer.valueOf(br.readLine());
            nVolume = nVolume2-nVolume1;
            nPack = Integer.valueOf(br.readLine());
            System.out.println("npack:" + nPack);
            for(int i=0; i<nPack; ++i)
            {
                value[i] = Integer.valueOf(br.readLine());
                System.out.println("value[i]" + value[i]);
                weight[i] = Integer.valueOf(br.readLine());
                System.out.println("weight[i]" + weight[i]);
            }
            for(int i=0; i<=nVolume; ++i)
                record[i] = INF;
            record[0] = 0;
            for(int i=0; i<nPack; ++i)
                for(int j=weight[i]; j<=nVolume; ++j)
                    if(record[j]>record[j-weight[i]]+value[i])
                        record[j] = record[j-weight[i]]+value[i];
            if(record[nVolume] == INF)
                System.out.println("This is impossible.\n");
            else
                System.out.println("The minimum amount of money in the piggy-bank is " + record[nVolume]);

        }

    }