杭电OJ——1114 Piggy-Bank（完全背包）

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.
   
   
   
   

 

Source

Central Europe 1999

 

Recommend

Eddy

 

     完全背包问题，不懂的可以去看看背包九讲！

     代码如下：

//完全没有想到，这道题目居然是一个完全背包问题！
/*
#include<iostream>
#include<algorithm>
using namespace std;

struct Coin
{
	int num,weight;
	float scale;
};

bool cmp(Coin a,Coin b)
{
	return a.scale>b.scale;
}

int main()
{
	Coin coin[100];
	int num,sum,i;
	cin>>num;
	while(num--)
	{
		int e,t,temp,a;
		int cases;
		cin>>e>>t>>cases;
		for(i=0;i<cases;i++)
		{
			cin>>coin[i].num>>coin[i].weight;
			coin[i].scale=(float)coin[i].weight/(float)coin[i].num;
		}
		sort(coin,coin+cases,cmp);
		temp=t-e;
		if(temp<0)
		{
			cout<<"This is impossible."<<endl;
			continue;
		}
		sum=0;
		for(i=0;i<cases;i++)
		{
			a=temp/coin[i].weight;
			temp=temp-coin[i].weight*a;
			sum=sum+coin[i].num*a;
		}
		//if(temp!=0)
        //cout<<"This is impossible."<<endl;
		//else
		printf("The minimum amount of money in the piggy-bank is %d.\n",sum);


	}
	return 0;
}
*/
/*
#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    int t,e,f,p[501],w[501],n,k[10001],fe;
    cin>>t;
    while(t--)
    {
        cin>>e>>f>>n;
        fe = f - e;
        for(int i = 0; i < n ; i++)
        {
           cin>>p[i]>>w[i];
        }
        k[0] = 0;
		//完全背包的初始化！
        for(int i = 1 ; i <= fe;i++)
        {
            k[i] = 50000000;
        }
        for(int i = 0 ; i < n ; i++)//选取硬币的种数
        {
            for(int m = w[i] ; m <= fe;m++)//重量
            {
                if(k[m-w[i]]+p[i]<k[m])//选取最小的
                k[m] = k[m-w[i]] + p[i];
            }
        }
        if(k[fe] == 50000000)
        printf("This is impossible.\n");
        else printf("The minimum amount of money in the piggy-bank is %d.\n",k[fe]);
    }
    return 0;
}
*/

#include<iostream>
using namespace std;
const int MAX=10050;

int f[MAX],w[MAX];//f数组记录硬币的面值，w数组记录硬币的重量
int k[MAX];//k数组用来记录最小取值

int main()
{
	int num,i;
	cin>>num;
	while(num--)
	{
		int e,t,temp;
		int cases;
		cin>>e>>t>>cases;
		temp=t-e;

		for(i=0;i<cases;i++)
		cin>>f[i]>>w[i];

		k[0]=0;//重量为0的钱币价值为0，即没有钱币

		for(i=1;i<=temp;i++)
		{
			k[i]=999999999;
		}

		for(i=0;i<cases;i++)
			for(int m=w[i];m<=temp;m++)
				if(k[m-w[i]]+f[i]<k[m])
					k[m]=k[m-w[i]]+f[i];

		if(k[temp]==999999999)
			printf("This is impossible.\n");
		else
			printf("The minimum amount of money in the piggy-bank is %d.\n",k[temp]);
	}
	//system("pause");
	return 0;

}