Hduoj1114【完全背包】

/*Piggy-Bank
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12661    Accepted Submission(s): 6409


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, P and 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 
*/ 
#include<stdio.h>
#define min(A,B)  (A)<(B)?(A):(B)
#define INF 1000000000
int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int a, b;
		scanf("%d%d", &a, &b);
		b -= a;//保存净重
		int n, f[10010], vo[550], va[550];
		scanf("%d", &n); 
		for(int i = 1; i <= n; i++)
		scanf("%d%d", &va[i], &vo[i]);
		f[0] = 0;
		for(int i = 1; i <= b; i++)//初始化
		f[i] = INF;
		for(int i = 1; i <= n; i++)
		{
			for(int j = vo[i]; j <= b; j++)
			{
				f[j] = min(f[j], f[j - vo[i]] + va[i]);//求出最小值最优解
			}
		} 
		if(f[b] == INF)
		printf("This is impossible.\n");
		else
		printf("The minimum amount of money in the piggy-bank is %d.\n", f[b]);
	} 
	return 0;
} 

题意：给出一个储蓄罐的净重，并且给出硬币的种类，每种硬币对应的有一个重量和价值，求储蓄罐中所能放下的最小价值，且要求最小价值的硬币重量刚好与储蓄罐的净重相等，其次每种硬币的数量是不限的。

思路：本题是一个完全背包，注意初始化时将f【0】 = 0，其它的全为无穷大，这是为了保证当有最小值的时候，其重量刚好等于净重，否则无解。