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!
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.
解题思路：已知空小猪罐重量与装满钱时的重量，给出每种币值与重量（每种可有无限），求出在这n种硬币中挑选出若干个硬币恰好装满小猪罐时得到的最小价值。理解了01背包恰好装满的情况，那么这个基础的完全背包恰好装满的问题也就变得简单了。完全背包恰好装满得到的最小价值的几个关键步骤：memset(dp,0x3f,sizeof(dp));dp[0]=0;dp[j]=min(dp[j],dp[j-w[i]]+v[i]);如果dp[W]==INF，说明不能恰好装满完全背包，否则可得到恰好装满时的最小价值。
AC代码：
#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
int T,E,F,N,W,v,w,dp[];
int main(){
while(cin>>T){
while(T--){
memset(dp,0x3f,sizeof(dp));dp[]=;//空罐时的最小价值为0
cin>>E>>F>>N;W=F-E;//完全背包的总重量
for(int i=;i<=N;i++){
cin>>v>>w;//v是币值,w是该种币的重量
for(int j=w;j<=W;j++)dp[j]=min(dp[j],dp[j-w]+v);
}
if(dp[W]!=INF)cout<<"The minimum amount of money in the piggy-bank is "<<dp[W]<<'.'<<endl;
else cout<<"This is impossible."<<endl;
}
}
return ;
}