Hdu1114--piggy-bank (full backpack variant)

Piggy-bank

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)

Total submission (s): 557 Accepted Submission (s): 304

Problem Descriptionbefore ACM can do anything, a budget must is prepared and the necessary financial support obtained. The main income for this action comes from irreversibly Bound money (IBM). The idea behind are simple. Whenever some ACM member have any small money, he takes all the coins and throws them into a piggy-bank. You know it is irreversible and the coins cannot be removed without breaking the pig. After a sufficiently long time, there should is enough cash in the Piggy-bank to pay everything that needs to be paid. But there was a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might the "the pig into Pieces" only "find out" that there are not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility was to weigh the piggy-bank, and try to guess what many coins is inside. Assume that we is able to determine the weight of the pig exactly and so we know the weights of all coins of a given CU Rrency. Then there are some minimum amount of money in the Piggy-bank so 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!

inputthe input consists of T test cases. The number of them (T) is given in the first line of the input file. Each test case is begins with a line containing the integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights is given in grams. No Pig would weigh more than ten kg, that means 1 <= E <= F <= 10000. On the second line of all test case, there is a integer number N (1 <= n <=) that gives the number of various Coins used in the given currency. Following this is exactly N lines, each specifying one coin type. These lines contain, integers each, pand w (1 <= P <= 50000, 1 <= w <=10000). P is the value of the coin in monetary units, and W is it's weight in grams.

Outputprint exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the Piggy-bank are X." where x is the minimum amount of Money that can is achieved using coins with the given total weight. If the weight cannot was 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 are 60.The minimum amount of money in the Piggy-bank are 100.This is Impossibl E.

Sourcecentral Europe 1999

Recommendeddy

Main topic:

A variety of coins into a piggy bank (unlimited number), each coin I have value Vi with weight Wi two properties, for the given weight y-x the smallest coin money n

Problem Solving Ideas:

This is a variant of the multi-pack (y-x)/wi per coin. But this question asks for the smallest value under the premise of fully filling the backpack.

The natural thought is to change the state transition equation of the classic knapsack problem and change Max to Min, but if you do so, you need to move your brains when initializing.

Consider the normal knapsack problem, if the requirement is exactly full state is, need to be except dp[0] all other elements are initialized to negative infinity, the final result if the negative infinity, indicating that no solution (negative infinity + finite number is still negative infinity, so that does not exactly fill the state).

But this topic requires a minimum value, so we need to initialize all elements other than dp[0] to positive Infinity , dp[0] to 0

This allows AC

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#defineINF 0X7FFFFFF#defineMAXN 10000using namespacestd;intdp[10010];intMain () {#ifndef Online_judge freopen ("data.in","R", stdin); #endifStd::ios::sync_with_stdio (false); Std::cin.tie (0); intT; intcc; inte,f; intN; CIN>>T; intp,w;  while(t--) {cin>>e>>F; CC=f-e; CIN>>N; dp[0]=0;  for(intI=1; i<=cc;i++) {Dp[i]=INF; }         for(intI=0; i<n;i++) {cin>>p>>W;  for(intj=w;j<=cc;j++) {Dp[j]=min (dp[j],dp[j-w]+p); }        }        if(dp[cc]==INF) {cout<<"This is impossible."<<Endl; }        Elsecout<<"The minimum amount of money in the Piggy-bank is"<<dp[cc]<<"."<<Endl; }}

Hdu1114--piggy-bank (full backpack variant)