HDU1114 完全背包变形

Piggy-Bank

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1902    Accepted Submission(s): 961

Problem Description

Before ACM can do anything, a budget must be prepared andthe necessary financial support obtained. The main income for this action comesfrom Irreversibly Bound Money (IBM). The idea behind is simple. Whenever someACM member has any small money, he takes all the coins and throws them into apiggy-bank. You know that this process is irreversible, the coins cannot beremoved without breaking the pig. After a sufficiently long time, there shouldbe 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 determinehow much money is inside. So we might break the pig into pieces only to findout that there is not enough money. Clearly, we want to avoid this unpleasantsituation. The only possibility is to weigh the piggy-bank and try to guess howmany coins are inside. Assume that we are able to determine the weight of thepig exactly and that we know the weights of all coins of a given currency. Thenthere 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 ofcash 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 aline containing two integers E and F. They indicate the weight of an empty pigand of the pig filled with coins. Both weights are given in grams. No pig willweigh more than 10 kg, thatmeans 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 ofvarious 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 inmonetary units, W is it's weight in grams.

 

 

Output

Print exactly one line of output for each test case. Theline must contain the sentence "The minimum amount of money in thepiggy-bank is X." where X is the minimum amount of money that can beachieved using coins with the given total weight. If the weight cannot bereached 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.

 

 

完全背包的变形,改成背包问题的描述就是,求出把背包装满可以获得的最小的价值.

 

#include<stdio.h>#include<string.h>#define N 505#define inf -2000000000inline int min (int a,int b){ return a>b?b:a;}int main(){ int cas,ori,i,j,n,p[N],w[N],dp[10005],total; scanf("%d",&cas); while (cas--) { scanf("%d%d",&ori,&total); total-=ori; scanf("%d",&n); for (i=1;i<=n;++i) { scanf("%d%d",&p[i],&w[i]); } for (i=1;i<=total;++i) dp[i]=inf; dp[0]=0; for (i=1;i<=n;++i) { for (j=w[i];j<=total;++j) { if (dp[j]<0) dp[j]=dp[j-w[i]]+p[i]; else { if (dp[j-w[i]]+p[i]>0) dp[j]=min(dp[j],dp[j-w[i]]+p[i]); } } } if (dp[total]>0) printf("The minimum amount of money in the piggy-bank is %d./n",dp[total]); else printf("This is impossible./n"); } return 0;}