动态规划—完全背包（装满）

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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.

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完全背包与01背包不同之处仅在于一个物品可以无数次选取（只要背包装得下），思路和01背包没多大区别，只是多了一层循环而已，找出要放几个。

状态转移方程：dp[i][j]=max(dp[i-1][j]，dp[i-1][j-k*w[i]]+k*v[i])，w[i]是重量，v[i]是价值。

当然也可以用一维数组优化，和01背包的一维数组做法几乎一样，不清楚的可以先看下我上一篇写的01背包https://www.cnblogs.com/hialin545/p/12257883.html，不过是内层的for循环变成了顺序的，

即：for(int j=w[i],j<=cap,j++)dp[j]=max(dp[j]，dp[j-w[i]]+v[i].

正序的话dp[j-w[i]]是当前第i下的，不是i-1，这样就是把第i件物品算进去了，建议自己拿笔模拟一下。

这道题目要求的是最小值，而且要判断是否装满，所以状态转移方程要修改一下：dp[j]=min(dp[j]，dp[j-w[i]]+v[i].

但是dp初始化时，要让dp[0]=0，其他的都dp[j]都为正无穷，关于背包装满的问题，具体可以看下这位大牛的博客：https://www.cnblogs.com/buddho/p/7867920.html

很清晰，我在这里看了以后就搞懂了。

然后贴代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;
int T,p[501],w[501],E,F,dp[10001],N;
int main()
{
  cin>>T;
  while(T--)
  {
    memset(dp,inf,sizeof(dp));
    dp[0]=0;
    cin>>E>>F;
    int cap=F-E;
    cin>>N;
    for(int i=1;i<=N;i++)cin>>p[i]>>w[i];
    for(int i=1;i<=N;i++)
    {
      for(int j=w[i];j<=cap;j++)
      {
        dp[j]=min(dp[j],dp[j-w[i]]+p[i]);
      }
    }
    if(dp[cap]==inf)cout<<"This is impossible."<<endl;
    else cout<<"The minimum amount of money in the piggy-bank is "<<dp[cap]<<"."<<endl;
  }
}