HDU1114--Piggy-Bank(完全背包变形)

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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 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 Central Europe 1999

Recommend Eddy

题目大意:

多种硬币放入一个存钱罐中(不限数目), 每种硬币 i 有价值 Vi 与重量  Wi 两个属性, 求给定重量 y-x 下最小的硬币钱数n

解题思路:

这是一个多重背包的变形[每种硬币最多有 (y-x)/Wi 个]. 但这道题目求的是在完全装满背包的前提下最小的价值.

自然想到的是将经典的背包问题的状态转移方程进行修改, 将max改为min, 但这样处理的话, 初始化的时候需要动一下脑筋

考虑正常的背包问题, 若要求的是恰好装满的状态是, 需要将除 dp[0] 之外的其他元素全部初始化为 负无穷, 最后的结果若为负无穷,说明无解(负无穷+有限数仍为负无穷, 这样就排除掉了不恰好装满的状态) .

但这道题目要求的是最小值, 所以我们需要将需 dp[0] 之外的其他元素全部初始化为 正无穷 , dp[0]初始化为0

如此即可AC

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define INF 0x7ffffff
#define MAXN 10000
using namespace std;
int dp[10010];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
    #endif
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t;
    int cc;
    int e,f;
    int n;
    cin>>t;
    int p,w;
    while(t--){
        cin>>e>>f;
        cc=f-e;
        cin>>n;
        dp[0]=0;
        for(int i=1;i<=cc;i++){
            dp[i]=INF;
        }
        for(int i=0;i<n;i++){
            cin>>p>>w;
            for(int j=w;j<=cc;j++){
                dp[j]=min(dp[j],dp[j-w]+p);
            }
        }
        if(dp[cc]==INF){
            cout<<"This is impossible."<<endl;
        }
        else cout<<"The minimum amount of money in the piggy-bank is "<<dp[cc]<<"."<<endl;
    }
}