hdu 1114 Piggy-Bank-优快云博客

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

题目描述：

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

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.

题目大意：

给你T组数据，每组数据第一行输入两个整数E、F，分别代表存钱罐没有装钱和装满钱时候的重量，接下来一行一个整数N，告诉你有N种钱币，每种钱币都有无数个，接下来的N行，每行两个整数P、W，分别代表该种钱币的价值和重量，问当存钱罐装满时，能装的最少价值的钱币是多少。如果不能装满，输出This is impossible.

题目分析：

这是一道完全背包的问题，不过这里是求最小价值，所以只需把dp[0]赋值0，其他都给+∞就可以了

AC代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
int T,emp,ful,n;
int p[10005],w[10005],dp[10005];
int main()
{
    cin>>T;
    while(T--)
    {
        cin>>emp>>ful;
        cin>>n;
        for(int i=1;i<=n;i++)
            {
                cin>>p[i]>>w[i];
            }
        for(int i=1;i<=ful-emp;i++)
            dp[i]=0xfffffff;
        dp[0]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=w[i];j<=ful-emp;j++)
                {
                    dp[j]=min(dp[j],dp[j-w[i]]+p[i]);
                }
        }
        if(dp[ful-emp]!=0xfffffff)
            printf("The minimum amount of money in the piggy-bank is %d.\n",dp[ful-emp]);
        else
            printf("This is impossible.\n");
    }
    return 0;
}