Mixing Milk
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 Source : Unknown Time limit : 3 sec Memory limit : 32 M
Submitted : 8651, Accepted : 3664
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.
Input
The first line contains 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.
The next M lines (Line 2 through M+1) each contain two integers, Pi and Ai. Pi (0 <= Pi <= 1,000) is price in cents that farmer i charges. Ai (0 <= Ai <= 2,000,000) is the amount of milk that farmer i can sell to Merry Milk Makers per day.
Output
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 Input
100 55 209 403 108 806 30
Sample Output
630

/*This Code is Submitted by aprilvkuo for Problem 1003 at 2013-01-01 17:35:42*/#include <stdio.h>#include <stdlib.h>int main(){    int m,a[5002],b[5002],tmp1,tmp,i,j;    long long n,sum=0;    scanf("%lld%d",&n,&m);    for(i=1; i<=m; i++)        scanf("%d%d",&a[i],&b[i]);    for(i=1; i<=m; i++)    {        for(j=i+1; j<=m; j++)        {            if(a[j]<a[i])            {                tmp=a[j],a[j]=a[i],a[i]=tmp;                tmp1=b[j],b[j]=b[i],b[i]=tmp1;            }        }        n-=b[i];        sum+=(b[i]*a[i]);        if(n<=0)        {            sum+=n*a[i];            break;        }    }    printf("%lld\n",sum);} 

其实不用多说了，题目要求得到买牛奶方案的最优算法。

可以直接用贪心，每次买价格最便宜的牛奶，直至采购到足够的牛奶量。

先根据牛奶价格将奶农排序，然后价格从低到高采购牛奶。