Piggy-Bank（完全背包）

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

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.

 

题意：题目变成我们熟悉的完全背包就好做了。现在有一个背包，我知道这个背包现在的重量，相当于背包容积已知，有n种货币，每种货币价值为p，重量为w，每个货币可以取无数个。询问当这些货币总质量刚好为f-e时的价值。这样就转换成了我们熟悉的背包问题。

dp[j]表示重量为j时背包里边的货币的价值，因为求的时最小钱数，因此我们需要将dp处初始成很大的数。之后初态就是dp[0]=0;

完全背包代码：

for(int i=0;i<=w;i++)//w是背包容积
            dp[i]=0;
        dp[0]=0;
        for(int i=0;i<n;i++)
            for(int j=v[i];j<=w;j++)
                dp[j]=max(dp[j],dp[j-v[i]]+value[i]);

 

AC代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 505
#define maxm 10005
#define INF 10000000
using namespace std;

int dp[maxm];
int e,f,n,p[maxn],w[maxn];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&e,&f);
        f=f-e;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&p[i],&w[i]);
        }
        for(int i=0;i<=f;i++)
            dp[i]=INF;
        dp[0]=0;
        for(int i=0;i<n;i++)
            for(int j=w[i];j<=f;j++)
                dp[j]=min(dp[j],dp[j-w[i]]+p[i]);
            if(dp[f]==INF)printf("This is impossible.\n");
            else printf("The minimum amount of money in the piggy-bank is %d.\n",dp[f]);
    }
    return 0;
}