HDU1114: Piggy-Bank
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.
完全揹包问题，与01揹包不同的是，完全揹包要求的就是揹包必须完全装满，例如本题的第一个例子，存钱罐不装钱重量是10，装完是110，所以里面的所装的前的重量就是100，不能多也不能少，并且，完全揹包有个特点，就是东西都是无限的，就像本题的货币，不限量，在动态规划过程中，dp需要从小到大遍历，具体分析在代码部分讲解
代码如下：
#include<iostream>
#include<cstring>
#include<stdio.h>
using namespace std;
int main()
{
int T;
int start,now; //声明存钱罐原先的重量和现在的重量
int N; //货币的种类
int value[509],weight[509],dp[10009];//分别代表货币的价值，货币的重量，动态规划揹包重量所能装的最小价值
int temp;//中间变量
cin>>T;
while(T--)
{
//输入
cin>>start>>now;
cin>>N;
for(int i=1;i<=N;i++){
cin>>value[i]>>weight[i];
}
//memset(dp,0x3f,sizeof(dp)指的是把dp这个数组的每一个字节都变成0x3f，而每个字节的最大值是0x3f，所以这行代码
代表的是把dp数组变得最大
memset(dp,0x3f,sizeof(dp));
dp[0]=0;//同时把dp[0]变成0，下面能起到一个关键的作用
for(int i=1;i<=N;i++){ //遍历每一种货币
for(int j=weight[i];j<=now-start;j++){
temp=value[i]+dp[j-weight[i]];
if(temp>0)
if(temp<dp[j])
dp[j]=temp;
}
}
/*“ temp=value[i]+dp[j-weight[i]];“，value[i]是当前货币的价值，这个货币的重量是weight[i]，
dp[j-weight[i]]的意义是当前存钱罐中钱的重量为 j ，dp[j-weight[i]]表示剩下的重量在之前所能代
表的最小价值
在上面的代码中初始化dp时，其他都是最大，唯独dp[0]=0，原因就是，假如有一块钱的货币重量为1，那么
dp[0]=0，dp[1]=value[1]+dp[0], dp[0]的用处便是可以在遍历过程中减少麻烦
而为什么要判断temp>0呢？开头也说过初始化dp，把它变成一个最大的数，只要这个数加上另一个大于零的数，
那么结果就会溢出，变成一个负数。这一步的所用就在于，假如当前遍历的货币是1，重量是2，且只有这一种
货币，那么遍历是从2开始的，当运算到dp[3]时，dp[3]=1+dp[1];算出的结果dp[3]等于一个最大数加上1，
溢出，变成一个最小数。
而 if(temp<dp[j])则是判断假如当前的货币，假如实际价值更小，则替换。因为结果是要输出存钱罐里最少有多少钱
*/
//假如最终dp[now-start]==dp[10008] ，则代表找不到能匹配的货币
if(dp[now-start]==dp[10008]){
cout<<"This is impossible."<<endl;
}
else
{
//输出结果
printf("The minimum amount of money in the piggy-bank is %d.\n",dp[now-start]);
}
}
}