Mixing Milk

Problem：

Mixing Milk

Since milk packaging is such a low margin business, it is important to keep the price of the raw product (milk) as low as possible. Help Merry Milk Makers get the milk they need in the cheapest possible manner.

The Merry Milk Makers company has several farmers from which they may buy milk, and each one has a (potentially) different price at which they sell to the milk packing plant. Moreover, as a cow can only produce so much milk a day, the farmers only have so much milk to sell per day. Each day, Merry Milk Makers can purchase an integral amount of milk from each farmer, less than or equal to the farmer's limit.

Given the Merry Milk Makers' daily requirement of milk, along with the cost per gallon and amount of available milk for each farmer, calculate the minimum amount of money that it takes to fulfill the Merry Milk Makers' requirements.

Note: The total milk produced per day by the farmers will be sufficient to meet the demands of the Merry Milk Makers.

PROGRAM NAME: milk

INPUT FORMAT

Line 1: Two integers, N and M.
The first value, N, (0 <= N <= 2,000,000) is the amount of milk that Merry Milk Makers' want per day. The second, M, (0 <= M <= 5,000) is the number of farmers that they may buy from.
Lines 2 through M+1: The next M lines each contain two integers, P_i and A_i.
P_i (0 <= P_i <= 1,000) is price in cents that farmer i charges.
A_i (0 <= Ai <= 2,000,000) is the amount of milk that farmer i can sell to Merry Milk Makers per day.

 

SAMPLE INPUT (file milk.in)

100 55 209 403 108 806 30

OUTPUT FORMAT

A single line with a single integer that is the minimum price that Merry Milk Makers can get their milk at for one day.

SAMPLE OUTPUT (file milk.out)

630

 

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My Answer：

/*ID:iby071PROG:milkLANG:C++*/# include<iostream># include<fstream>using namespace std;struct node{int price;int amount;};node *searchlow(node *a,int size)//用冒泡法把数组中的待搜索项中的最小值移到最后，并返回它的地址（注意，返回值是指针）{node tmp;for(int i=0;i<size;++i)if(a[i+1].price>a[i].price){tmp=a[i];a[i]=a[i+1];a[i+1]=tmp;}return &a[size-1];}int main(){ifstream fin("milk.in");ofstream fout("milk.out");int total,farmer,money=0,i=0;node *p;fin>>total>>farmer;node *data=new node[farmer];for(int j=0;j<farmer;++j) fin>>data[j].price>>data[j].amount;while(total>0){p=searchlow(data,farmer-i);//每次搜索1个最小项，并把它冒泡到了最后。下次搜索至它的前一个元素就好。if(total>=p->amount){money+=p->price*p->amount; total-=p->amount;}else{ money+=p->price*total;total=0;}//处理最后一个供奶商，他拥有的奶总量如果大于所需量了，不能再直接price*amount，而应该price*total（剩余）++i;}fout<<money<<endl;return 0;}

My Gain:

在这个程序之前写了一个用到快排的程序。
所以当时首先想到的是对数组全部排序依次，再依次取出。
结果提交以后，报了时间过长的错误。
然后想到，反正只需要找到几个price最便宜的供奶商，使得它们的牛奶amount之和达到total就足够了。
完全不需要对整个数组进行排序。
所以就想到，每一次都冒泡找到最小的，又把它放在所有参与冒泡元素的最后。
下一次冒泡的时候冒到它的前一个就为止，就找到了次小的。
以此类推，直到买够了total的牛奶。

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