HDU 1114 Piggy-Bank

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16743    Accepted Submission(s): 8452

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.

 

完全背包问题。题意为给定一个容器的重量和盛满以后的重量。然后给定几种类型的物品的价值和重量，问在盛满的情况下，可以产生的最少的价值。如果无法到达的话，那么就输出impossible

这个题目的特殊之处在于一定要把背包填满。而且需要最少值。所以讲除了dp[0]以外的都赋值为正无穷。

根据完全背包的转移方程有：dp[j] = min(dp[j], dp[j - w[i]] + v[i]);

代码如下；

/*************************************************************************
	> File Name: Piggy_Bank.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Sat 24 Oct 2015 12:10:03 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#define INF 1<<29
using namespace std;
int T, E, F;
int n;
int dp[10010];
int w[510];
int v[510];
int main(void){
    cin >> T;
    while(T --){
        cin >> E >> F;
        cin >> n;
        for(int i = 0; i < n; i ++){
            scanf("%d%d", &v[i], &w[i]);
        }
        for(int j = 1; j <= F - E; j ++)
            dp[j] = INF;
        dp[0] = 0;
        for(int i = 0; i < n; i ++){
            for(int j = w[i]; j <= F - E; j ++){
                dp[j] = min(dp[j], dp[j - w[i]] + v[i]);
            }
        }
        if(dp[F - E] != INF)
            printf("The minimum amount of money in the piggy-bank is %d.\n", dp[F - E]);
        else
            printf("This is impossible.\n");
    }
    return 0;
}