Piggy-Bank （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.

 

source

 

就是说输入测试数据数

输入空的存钱罐的重量以及满了的重量

输入硬币种类数t

以下t行输入t种硬币每个硬币的重量以及面值

求此情况下该存钱罐至少能存的钱数，无解就输出无解

 

和硬币那道题有相似之处，那道题是给出几种钱币，每种钱币的数量无限，算资金一定的时候的钱币的组合数

这道题是给出几种钱币的价值和重量，每种钱币的数量无限，算背包容量一定的情况下所存的最少钱数，也可以拓展到最多钱数，把min改为max就好了

动态和背包好像

#include <stdio.h>
#define INF  0x3f3f3f3f
#include <algorithm>
using namespace std;

int t, w[], val[], dp[], w0, w1; //w0存空存钱罐重量，w1存装满的存钱罐重量，dp[i]代表容量为i的时候所能存的最少钱数，t钱币种类数
void work()
{
    for(int i = ; i <= w1 - w0; i++)
        dp[i] = INF; // 初始标记为无穷，如果计算完之后任然为无穷的话说明无解，小于无穷则输出结果
    dp[] = ;//容量为0所能存的资金也是0
    for(int i = ; i < t; i++)
    {
        for(int j = w[i]; j <= w1 - w0; j++)
            dp[j] = min(dp[j], dp[j-w[i]] + val[i]);//算出只放i种钱币，其中第i种钱币放（0个——所能放的最多数量）的时候存钱罐里所存的的最少资金
    }
}
int main()
{

    int n;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%d%d", &w0, &w1);
        scanf("%d", &t);
        for(int i = ; i < t; i++)
        {
            scanf("%d%d", &val[i], &w[i]);
        }
        work();
        if(dp[w1 - w0] < INF)
            printf("The minimum amount of money in the piggy-bank is %d.\n", dp[w1 - w0]);
        else
            printf("This is impossible.\n");
    }
    return ;
}