3018

3018

Problem R

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 58   Accepted Submission(s) : 27

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. <br><br>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! <br>

 

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. <br>

 

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.". <br>

 

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：
    给出小猪钱罐的重量和装满钱后的重量，然后是几组数据，每组数据包括每种钱币的价值与重量
    要求出重量最少能装满钱罐时的最大价值
思路：
     很无奈的是，读了好几遍题之后仍然看不出它到底想做啥。更是完全没有想到跟完全背包有什么关系。但后来被点醒之后，问题就变得很明朗了。
     完全背包，与0-1背包不同的是第二次遍历的顺序，倒过来就行，求最小值就改成MIN，其他的没什么变化。
     另外要注意初始化的值。
     因为题目要求恰好装满，而且要求的值要尽量小，所以将dp[0]设为0，其余设为无穷大（如果没有要求恰好装满的话，全0）。
 

AC代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cctype>
#include<stdio.h>
#define min(a,b)(a<b?a:b)
#define max(a,b)(a>b?a:b)
#define INF 0x3f3f3f3f
typedef long long ll;
#define N 501
int p[N],w[N],dp[10001];


int main()
{
   int T,E,F,n,i,j;
   scanf("%d",&T);
   while(T--)
   {
       scanf("%d%d",&E,&F);
       int V=F-E;  /// 背包容量为装满时减去为空时
       scanf("%d",&n);
       for(i=0;i<n;i++)
       scanf("%d%d",&p[i],&w[i]);


       for(i=0;i<=V;i++)
           dp[i]=INF;  ///初始化～为最大值


       dp[0]=0;
       for(i=0;i<n;i++)
       {
           for(j=w[i];j<=V;j++)  ///简单的完全被包模版～
           {
               dp[j]=min(dp[j],dp[j-w[i]]+p[i]);
           }
       }


       if(dp[V]==INF)
        printf("This is impossible.\n");
       else
        printf("The minimum amount of money in the piggy-bank is %d.\n",dp[V]);
   }
}