HDU-1114 完全背包+恰好装满问题

B - Piggy-Bank

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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.

 

 

非常典型的完全背包以及恰好装满问题。

首先，要实现完全背包，循环背包容量的时候，即从0 到 最大容量，模拟一个矩阵数表也看得出，这样只要背包存得下，即从前一个f[j-v[i]]状态再加当前物品，等该层循环完毕，单种物品已经尽可能多的存在了背包当中。。。这是我对完全背包的理解。

 

然后是恰好装满问题，我一开始想的是，判断f[max]与f[max-1]是否相等，不过感觉略有些虚，好像对也好像不对。。。。然后网上大神给的标解是，将f[0]=0,其他值，如果是求背包最大，则设置为-inf（无穷小），如果是求背包最小，则设置为inf。。。这样想一下也很明晰，唯有f[0+某商品]才真正是背包容量，其他的无穷大和无穷小的状态都是虚的。。。针对本体来说，如果最终f[max]仍旧是无穷大，那说明根本凑不齐恰好装满这种状态。

 

然后还有一点心得是：本题是求背包最小值，我有点定向思维，写多了求最大值的背包，其实求最小值也是一样的写法，我目前的思路就是，将f[0]赋值为0，其他的都赋值为无穷大。然后在遍历过程中，取min值。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int f[10005];
int w[505],p[505];
int max(int x,int y)
{
    if (x>y) return y;
    return x;
}
int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int em,fu;
        scanf("%d %d",&em,&fu);
        int wei=fu-em;
        int n;
        scanf("%d",&n);
        int i,j,k;
        for (i=0;i<n;i++)
        {
            scanf("%d %d",&p[i],&w[i]);
        }
        for (i=1;i<=wei;i++) f[i]=1e9;
        f[0]=0;
        for (i=0;i<n;i++)
        {
            for (j=0;j<=wei;j++)  //从小到大遍历背包容量，完全背包
            {
                if (j-w[i]>=0)
                {
                    f[j]=min(f[j],f[j-w[i]]+p[i]);
                }
            }
        }
        if (f[wei]==1e9) puts("This is impossible.");//若满足条件，说明根本不能恰好装满。
        else
        printf("The minimum amount of money in the piggy-bank is %d.
",f[wei]);
    }
    return 0;
}