Piggy-Bank----HDU_1114----完全背包（变式）

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1114

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5033    Accepted Submission(s): 2479

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.
  

 

Source

Central Europe 1999

先分析题目的意思。这个题目的意思是有一只存钱的小猪灌，告你它空着的时候的重量，以及满着的时候的重量。然后有一些钱币，每种钱币都有自己的价值和重量。每种钱币可以使用任意的次数。现在题目的要求就是要你把这个储钱罐装满，但是里面装的钱币的价值要是最少的。这就和我们以前做过的完全背包问题相反了。怎么做呢？很简单。就是状态转移方程f[j]在赋初值的时候全部赋成正无穷大，表示在j背包容量下什么都不装所能达到的最小价值为正无穷，也就是一个无效的背包。但是f[0]要赋作0，因为在0背包容量下你什么都不能放进去，则最小的当然是0，是一个有效的状态。而整个背包的总容量就是用满重量的容量减去空的时候的容量。再根据完全背包的算法求出装满情况下最少的价值。如果最后f[V]为无效状态的话，则代表，这个背包根本就不能装满，那说明在所给的条件下是无法满足要求的，输出“This is impossible.”。如果不是，则说明存在最小值，则输出。完全背包思想详见：http://love-oriented.com/pack/P02.html

#include<iostream>
using namespace std;

#define MAX 10010
#define INF 999999999
int f[MAX];   //f[j]表示的是在j容量下装满时的最小价值，若为无穷大，则表示无法装满，状态无效 
int value[MAX];
int weight[MAX];

int min(int a,int b)
{
	if(a<b)
		return a;
	return b;
}


int solve(int n,int V)
{
	int i,v;
	for(i=1;i<=n;i++)  //采用的是完全背包算法 
	{
		for(v=weight[i];v<=V;v++)
		{
			f[v] = min(f[v],f[v-weight[i]]+value[i]);  //这就是常规题型不同的地方，是在有效的情况下求最小 
		}
	}

	return f[V];
}


int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int pig,fullpig;
		scanf("%d%d",&pig,&fullpig);
		int n;
		scanf("%d",&n);
		int i;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&value[i],&weight[i]);
		}
		for(i=1;i<=fullpig-pig;i++) //初始化，很重要 
			f[i] = INF;
		f[0] = 0;
		int ans = solve(n,fullpig-pig);

		if(ans >= INF)   //如果最大容量是一个无效的状态，也就是无法装满的情况 
			printf("This is impossible.\n");
		else
			printf("The minimum amount of money in the piggy-bank is %d.\n",ans);
	}

	return 0;
}