文章目錄

• • Problem Description
  • 題意：
  • 題解：
  • 代碼：

hdu-1114

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.

題意：

E和F分別表示空盒子和裝滿物品的盒子的質(zhì)量，然后給你N件物品，每件物品的價(jià)值是P，質(zhì)量是W，問(wèn)裝滿這個(gè)盒子最少用總價(jià)值為多少的物品
分析樣例：

10 110 2 1 1 30 50

要裝的物品質(zhì)量為110-10=100
裝兩個(gè)物品2（即價(jià)值30，質(zhì)量50的這個(gè)物品），即可使得盒子裝滿，且總價(jià)值最低

題解：

物品有價(jià)值有質(zhì)量，但是沒(méi)有數(shù)量，沒(méi)錯(cuò)就是完全背包問(wèn)題
然后套上完全背包的板子
背包問(wèn)題講解
因?yàn)槭乔笞钚r(jià)值，記得將max改成min

代碼：

懶得敲代碼了，代碼來(lái)自

#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int INF = 500500; int dp[INF],val[INF],we[INF]; int main(void) {int t,n,i,j,m,st,ed,tp;cin>>t;while(t--){cin>>st>>ed;cin>>n;tp=ed-st;for(i=0;i<n;i++) cin>>val[i]>>we[i];memset(dp,INF,sizeof(dp));dp[0]=0;for(i=0;i<n;i++){for(j=we[i];j<=tp;j++){dp[j]=min(dp[j],dp[j-we[i]]+val[i]);}}if(dp[tp]!=dp[500100]) printf("The minimum amount of money in the piggy-bank is %d.\n",dp[tp]);else printf("This is impossible.\n");}return 0; }

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