동적 계획 - 전체 가방(가득 채움)

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.

완전 배낭과 01배낭의 다른 점은 단지 하나의 물품을 여러 번 선택할 수 있기 때문(등만 포장할 수 있다면)이다. 사고방식은 01배낭과 크게 다르지 않고 한 층의 순환이 많을 뿐 몇 개를 넣어야 할지 찾아낸다.
상태 이동 방정식: dp[i][j]=max(dp[i-1][j], dp[i-1][j-k*w[i]]]+k*v[i]), w[i]는 무게, v[i]는 가치.
물론 1차원 그룹으로 최적화할 수도 있습니다. 01배낭의 1차원 그룹과 방법은 거의 같습니다. 잘 모르겠으면 제가 지난번에 쓴 01배낭을 먼저 보셔도 됩니다.https://www.cnblogs.com/hialin545/p/12257883.html, 내부 for 순환이 순서로 바뀌었을 뿐,
즉, for(int j=w[i], j<=cap, j++)dp[j]=max(dp[j], dp[j-w[i]]+v[i].
정서적으로 보면 dp[j-w[i]는 현재 i에서 나온 것이지 i-1이 아니다. 그러면 i번째 물품을 계산한 것이다. 스스로 펜을 가지고 시뮬레이션을 해보자.
이 문제는 최소값을 요구하고 가득 찼는지 판단해야 하기 때문에 상태 이동 방정식을 수정해야 한다. dp[j]=min(dp[j], dp[j-w[i]]+v[i].
그러나 dp를 초기화할 때 dp[0]=0을 만들어야 한다. 다른 것은 모두 dp[j]가 무궁무진하다. 가방이 가득 찬 문제에 대해 구체적으로 이 소의 블로그를 볼 수 있다.https://www.cnblogs.com/buddho/p/7867920.html
분명히 나는 여기서 본 후에 이해했다.
그런 다음 코드를 붙여 넣습니다.

 1 #include
 2 #include
 3 #include
 4 #include
 5 #define inf 0x3f3f3f3f
 6 using namespace std;
 7 int T,p[501],w[501],E,F,dp[10001],N;
 8 int main()
 9 {
10   cin>>T;
11   while(T--)
12   {
13     memset(dp,inf,sizeof(dp));
14     dp[0]=0;
15     cin>>E>>F;
16     int cap=F-E;
17     cin>>N;
18     for(int i=1;i<=N;i++)cin>>p[i]>>w[i];
19     for(int i=1;i<=N;i++)
20     {
21       for(int j=w[i];j<=cap;j++)
22       {
23         dp[j]=min(dp[j],dp[j-w[i]]+p[i]);
24       }
25     }
26     if(dp[cap]==inf)cout<<"This is impossible."<<endl;
27     else cout<<"The minimum amount of money in the piggy-bank is "<"."<<endl;
28   }
29 }