HDU 1114：Piggy-Bank

Piggy-Bank

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27918    Accepted Submission(s): 14096

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

310 11021 130 5010 11021 150 301 6210 320 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

题意：

给定物品种类和所购物品质量，又给定相应的物品的价格和重量，求花费最小的花费，恰好达到所需的重量。

解题思路：

不难看出这是一个完全背包问题而且是恰好装满的背包问题。但是，不同点是，该题是花费最小的钱，得到最大的质量，与普通完全背包恰好相反。既然让求最小的花费，我们不妨把dp初值赋值为INF，dp[0]=0;

动态转移方程：dp[j]=min(dp[j],dp[j-w[i]]+v[i]);//j动态转移方程下的所需质量，w[i],v[i]分别表示物品质量和价格。

源代码：

#include<cstdio>#include<cstring>int min(int a,int b){return a<b?a:b;}const int N=10010;const int INF=0x3f3f3f3f;int main(){    int t;    scanf("%d",&t);    while(t--)    {        int w1,w2,n;        int w[N],v[N],dp[N];        scanf("%d%d",&w1,&w2);        int W=w2-w1;        memset(dp,INF,sizeof(dp));//与普通恰好完全背包恰好相反，因为这是求最小花费        dp[0]=0;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d%d",&v[i],&w[i]);        }        for(int i=0;i<n;i++)        {            for(int j=w[i];j<=W;j++)            {                dp[j]=min(dp[j],dp[j-w[i]]+v[i]);//dp[j]这个方程下的最小花费            }        }        if(dp[W]!=INF)//能恰好装满            printf("The minimum amount of money in the piggy-bank is %d.\n",dp[W]);        else            printf("This is impossible.\n");    }}