HDU 1114 Piggy-Bank 【完全背包】
2015-09-12 16:57
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Piggy-Bank
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16443 Accepted Submission(s): 8299
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 hasany 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
[align=left][/align]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 totalweight. If the weight cannot be reached exactly, print a line "This is impossible.".
Sample Input
[align=left][/align]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嗯,题目大意就是说,小猪存钱罐,给出空的和满的重量,然后给出一些硬币的面额和重量,问小猪存钱罐里最少有多少钱。输入数据时首先是测试数据组数,之后是空的和满的重量,然后是硬币的种类数,下面就是硬币的面额和重量了。这个也相当于完全背包恰好装满的情况所以f 数组首先初始化为一个较大的数,这里要求的是最少的钱数
#include <iostream> #include<cstdio> #include<cstring> #define maxn 550 #include<algorithm> using namespace std; int f[100100],w[maxn],p[maxn]; int min(int a,int b) { return a<b?a:b; } int main() { int t,E,F,n; scanf("%d",&t); while(t--) { scanf("%d%d",&E,&F); scanf("%d",&n); for(int i=0;i<n;++i) scanf("%d%d",&p[i],&w[i]); memset(f,0,sizeof(f)); F=F-E; for(int i=1;i<=F;++i) f[i]=999999;//我用1000就WA了,所以尽量取大一点 for(int i=0;i<n;++i) { for(int j=w[i];j<=F;j++) f[j]=min(f[j],f[j-w[i]]+p[i]); } if(f[F]==999999) printf("This is impossible.\n"); else printf("The minimum amount of money in the piggy-bank is %d.\n",f[F]); } return 0; }
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