Piggy-Bank
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 6 Accepted Submission(s) : 6
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
PKU
属于完全背包问题，只是把最大值改成求最小值。
状态转移方程：DP[i]=min(DP[i],DP[i-Wight]+Value);
时间复杂度：O（N*N）;
1 #include2 #include 3 #define INF 100000000 4 5 int DP[10100];/*DP[j]代表的是,容量为j的背包所能存放的价值*/ 6 put() 7 { 8 int i; 9 putchar(10);10 for(i=0;i<=5;i++)11 printf("%d ",DP[i]);12 putchar(10);13 }14 int main()15 {16 int Wight1,Wight2,Value,Wight,T,n,i,j;17 scanf("%d",&T); /*输入有T组测试样例*/18 while(T--) /*循环操作T次*/19 {20 scanf("%d%d%d",&Wight1,&Wight2,&n); /*每一组测试样例，先输入空罐和满罐的重量，还有n中硬币*/21 for(i=1;i<=Wight2-Wight1;i++) /*初始化DP[]，因为题目要求背包需要全部装满，无法确定一个初始值*/22 DP[i]=INF; /*并且题目所要求的是最小值，初始化为一个极大值 */23 DP[0]=0;24 while(n--)25 {26 scanf("%d%d",&Value,&Wight); /*输入硬币的价值和硬币的重量*/27 for(i=Wight;i<=Wight2-Wight1;i++) /*因为一个硬币的数量是无限的，所以要从递增顺序开始循环*/28 if(DP[i]>DP[i-Wight]+Value) /*容量为i的背包，与之前的比较，选择最小值存储在DP[i]中*/29 DP[i]=DP[i-Wight]+Value;30 put();31 }32 if(DP[Wight2-Wight1]==INF) /*如果目标重量的背包中的值没有被替换，说明不存在能是背包存满的情况*/33 printf("This is impossible.\n");34 else35 printf("The minimum amount of money in the piggy-bank is %d.\n",DP[Wight2-Wight1]);36 }37 return 0;38 }