HIT1003 Mixing Milk

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本文介绍了一个算法案例，针对MerryMilkMakers公司如何从多个供应商中以最低成本获取所需牛奶的问题，通过输入每日需求量及各供应商的价格与供应量，采用排序与累积求和的方法实现了最小化成本的目标。

Description
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 5
5 20
9 40
3 10
8 80
6 30
Sample Output
630
Solution

#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <stack>
#include <string>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

int n,m;

struct Node{
    int p;
    int a;
}f[5005];

bool cmp(Node a,Node b){
    if (a.p < b.p)return 1; else return 0;
}

int main(){
    int cnt=0;
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;++i){
        scanf("%d %d",&f[i].p,&f[i].a);
    }
    sort(f+1,f+1+m,cmp);
    for (int i=1;i<=m && n > 0;++i){
        if ((f[i].a) <= n){
            n-=(f[i].a);
            cnt +=   (f[i].a)*(f[i].p);
        }
        else{
            cnt += n * f[i].p;
            n = 0;
        }
    }
    printf("%d\n",cnt);
    return 0;
}