Piggy-Bank（最小完全背包问题）

Piggy-Bank

http://acm.hdu.edu.cn/showproblem.php?pid=1114

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 6

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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 

题目大意：

就是给定银行的最大重量，以及给出每个人的钱数及重量，求能恰好装满银行的最小钱数，如果不能恰好装满，则输出不可能

求最小值我参考的背包九讲，下面就是截取的背包九讲的内容：

初始化的细节问题

 

我们看到的求最优解的背包问题题目中，

事实上有两种不太相同的问法。

有的题

目要求“恰好装满背包”时的最优解，有的题目则并没有要求必须把背包装满。

一种区别这两种问法的实现方法是在初始化的时候有所不同。

 

如果是第一种问法，要求恰好装满背包，那么在初始化时除了

f[0]

为

0

其它

f[1..V]

均设为

-

∞，这样就可以保证最终得到的

f[N]

是一种恰好装满背包的最

优解。

 

如果并没有要求必须把背包装满，而是只希望价格尽量大，初始化时应该将

f[0..V]

全部设为

0

。

 

为什么呢？可以这样理解：

初始化的

f

数组事实上就是在没有任何物品可以放入

背包时的合法状态。

如果要求背包恰好装满，

那么此时只有容量为

0

的背包可能

被价值为

0

的

nothing“恰好装满”，其它容量的背包均没有合法的解，属于未

定义的状态，它们的值就都应该是

-

∞了。如果背包并非必须被装满，那么任何

容量的背包都有一个合法解“什么都不装”，这个解的价值为

0

，所以初始时状

态的值也就全部为

0

了。

初始化的细节问题

我们看到的求最优解的背包问题题目中，事实上有两种不太相同的问法。有的题目要求“恰好装满背包”时的最优解，有的题目则并没有要求必须把背包装满。一种区别这两种问法的实现方法是在初始化的时候有所不同。

如果是第一种问法，要求恰好装满背包，那么在初始化时除了f[0]为0其它f[1..V]均设为-∞，这样就可以保证最终得到的f[N]是一种恰好装满背包的最优解。

如果并没有要求必须把背包装满，而是只希望价格尽量大，初始化时应该将f[0..V]全部设为0。

为什么呢？可以这样理解：初始化的f数组事实上就是在没有任何物品可以放入背包时的合法状态。如果要求背包恰好装满，那么此时只有容量为0的背包可能被价值为0的nothing“恰好装满”，其它容量的背包均没有合法的解，属于未定义的状态，它们的值就都应该是-∞了。如果背包并非必须被装满，那么任何容量的背包都有一个合法解“什么都不装”，这个解的价值为0，所以初始时状态的值也就全部为0了。

这个小技巧完全可以推广到其它类型的背包问题，后面也就不再对进行状态转移之前的初始化进行讲解。

代码实现：

#include
#define MAX 2000000000//代表无穷大（注意，足够大就好，不要太大，以免出现溢出的危险导致结果混乱）
#define N 10000
int p[N],dp[N],v[N];
void set(int dp[],int V)//赋初值
{
    int i;
    dp[0]=0;
    for(i=1;i<=V;i++)
    {
        dp[i]=MAX;
    }
}
int main()
{
    int t,V,v1,v2,n,i,j;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            scanf("%d%d",&v1,&v2);
            V=v2-v1;
            scanf("%d",&n);
            for(i=0;i