hdu1114

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10149    Accepted Submission(s): 5113

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.

 
题意：
给出一个储钱罐，存钱罐本身有重量，给出几种钱的价值和重量，比如第一组数据
10     110---表示存钱罐的质量为10 存满质量为110所以钱的重量为100；
2-----两组数据；
1  1和30   50  分别为两种货币的价值和重量
分析一下如果全装第一组，价值为100，全装第二组为60，混装价值还是100，要求选取最小值，所以得到的结果为60；
做题思路：
一看这题就是个背包问题，很明显的完全背包，所以按完全背包的思路解但还有要注意的，题中要求求最小值，所以你初始化数组就不可以按完全背包初始为0，而是无穷大
而且对于背包的算法还要取两个最小值；
代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
#define inf 99999999
int dp[10005],w[501],v[501];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int m,n,k;
        cin>>m>>n;
        cin>>k;
        for(int i=0;i<k;i++)
        {
            cin>>v[i]>>w[i];
        }
        int v1=n-m;
        fill(dp,dp+v1+1,inf);
        dp[0]=0;
        for(int i=0;i<k;i++)
            for(int j=w[i];j<=v1;j++)
            {
                dp[j]= min(dp[ j - w[i] ]+v[i],dp[j] );//求最小值放入dp中
            }
        if(dp[v1]==inf)
            cout<<"This is impossible."<<endl;
        else
            cout<<"The minimum amount of money in the piggy-bank is "<<dp[v1]<<'.'<<endl;
    }
    return 0;
}