POJ1384 Piggy-Bank （完全背包）



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

310 11021 130 5010 11021 150 301 6210 320 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.

改过那么一点点的完全背包问题。与之前描述的01背包问题的不同在于，01背包的每个物品只能取1件，而完全背包的物品都有无数件。看这个题解之前建议大家去看看我的01背包的题解。里面在讲原地打滚的搜索方式时提到过，如果从左往右搜会导致一个物品背加多次，出现错误。但换过来想想，这不恰恰是完全背包的意思吗？这不恰恰是完全背包需要的吗？

不过要注意，这里是已知填充的重量，求花费的最小代价，程序会有两个地方不太一样：

1.初始化时正常的会把f初始化为0，但这里要初始化成一个很大的数

2.正常的是f[j]=max(f[j],f[j-w[i]]+p[i]),但这里是f[j]=min(...)

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;int main(){    int ii,t,e,c[502],w[502],f[10050],n,m,i,j,f0;   scanf("%d",&t);   for (ii=0;ii<=t-1;ii++)   {       memset(c,0,sizeof(c));       memset(w,0,sizeof(0));       memset(f,0,sizeof(f));       scanf("%d%d",&e,&f0);       m=f0-e;       scanf("%d",&n);       for (i=0;i<=n-1;i++)       {           scanf("%d%d",&c[i],&w[i]);       }       for (i=0;i<=m;i++)       {           f[i]=1000000;       }       f[0]=0;       for (i=0;i<=n-1;i++)       {           for (j=w[i];j<=m;j++)           {               f[j]=min(f[j],f[j-w[i]]+c[i]);           }       }       if(f[m]==1000000)       {           printf("This is impossible.\n");       }       else       {           printf("The minimum amount of money in the piggy-bank is %d.\n",f[m]);       }   }   return 0;}