HDU 1114 Piggy-Bank

Posted 2020-10-10 0一叶0知秋0

HDU 1114 Piggy-Bank

kuangbin 专题十二：  F题

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27652    Accepted Submission(s): 13976

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.

题意：
d.给出一个存钱罐的容量，给出n种硬币的价值p和重量w（注意：每种硬币可无限取）

1.如果存钱罐能够正好塞满，输出塞满存钱罐需要的最少硬币的价值。

2.若不能完全塞满，则输出impossible。
思路：
s.每种物品可以放无限多次。所以为完全背包问题。此题是求最小值，为完全背包的变形。

注意初始化 dp[ 0 ]=0;

for i=1..N

    for v=0..V

        f[v]=max{f[v],f[v-cost]+weight}

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std ;  

#define maxn 20000000
int dp[maxn] ; 

struct node {
    int p ; 
    int w ; 
};
node num[maxn] ; 
int main(){
    int t ; 
    int n ; 
    int v ; 
    int e , f ; 
    
    
    scanf("%d" , &t) ; 
    while(t--){
        scanf("%d %d" , &e , &f) ; 
        v = f - e ;  // 背板实际容量  
        scanf("%d" , &n) ; 
        for(int i=1 ; i<=n ; i++){
            scanf("%d%d" , &num[i].p , &num[i].w) ;     
        }
        for(int i=1 ; i<= v ; i++){
            dp[i] = 10000000 ; 
            // 因为要求  正好装满背包的最小价值  所以初始化 dp 数组为相反状态  
        }
        dp[0] = 0 ; 
        for(int i=1 ; i<=n ; i++){ 
            for(int j = num[i].w ; j <= v ; j++ ){
                dp[j] = min(dp[j] , dp[j-num[i].w] + num[i].p) ; 
            }
        }
        if(dp[v] == 10000000)// 如果 所有硬币没有正好装满背包  则dp[v] 不会有任何变化  
            printf("This is impossible.\n") ;
        else
            printf("The minimum amount of money in the piggy-bank is %d.\n" , dp[v]) ; 
    }
    
    return 0 ; 
}