【HDU】hdu 1114 （完全背包）

 

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.

 

 

这是一个动态规划 ，完全背包问题，计算最小的钱数:

 

 

/****************************************************************
   完全背包注意他的循环方向，和01背包第二个循环正好相反
   数组minmon【i】表示质量为i的时候取得钱的最小值
   那么递推关系式
   if (minmon[j + mon[ i ].wei]==0)minmon[j + mon[ i ].wei] =minmon[j]+mon[i].num;
   else minmon[j + mon[ i ].wei] = min(minmon[j]+mon[i].num,minmon[j + mon[ i ].wei]);
    最后访问看质量为钱的总质量是否访问过来确定时候可以达到
****************************************************************/
#include <iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
int empwe,fullwe,number;
int minmon[10005];
int vis[10005];
struct money
{
    int num;
    int wei;
};
int main()
{
    int test;
    cin >> test;
    while ( test > 0 )
    {
        test--;
        money mon[505];
        cin >> empwe >> fullwe >> number;
        int summon = fullwe - empwe;///计算出钱的质量
        for ( int i = 0; i < number; i++)
        {
            cin >> mon[i].num >> mon[i].wei;
        }
        memset(vis,0,sizeof(vis));
        memset(minmon,0,sizeof(minmon));
        vis[0] = 1;
        for ( int i = 0; i < number; i++ )
        {
            for ( int j = 0; j <= summon - mon[i].wei ; j++)
            {
                if ( vis [j])
                {
                    if (minmon[j + mon[ i ].wei]==0)
                        minmon[j + mon[ i ].wei] =minmon[j]+mon[i].num;///如果这里的最小值是0 的话就不能去最小值；
                    else
                        minmon[j + mon[ i ].wei] = min(minmon[j]+mon[i].num,minmon[j + mon[ i ].wei]);
                    vis[j + mon[ i ].wei] = 1;
                }
            }
        }
        if( vis[summon] ) ///如果这个质量没有访问过
            printf("The minimum amount of money in the piggy-bank is %d.\n",minmon[summon]);
        else cout << "This is impossible." << endl;

    }
}