USACO SEC.1.3 No.1 Mixing Milk

题意：需要收购总数为N的牛奶，现在有M个牛奶供应商（总量足够），给出总数和单价，求最小的花销。

核心：基本的贪心解法，按单价排序逐个选取。

目的在于熟悉基本的贪心法的基本方法和思路

/*
ID: lsswxr1
PROG: milk
LANG: C++
*/
#include <iostream>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <fstream>
using namespace std;

///宏定义
const int  INF = 1000000000;
const int MAXN = 5015;
const int maxn = MAXN;
///全局变量 和 函数

#define USACO
#ifdef USACO
#define cin fin
#define cout fout
#endif
//////////////////////////////////////////////////////////////////////////
int N, M;
struct farm
{
    int price, total;
    bool operator < (const farm& t) const
    {
        return price < t.price;
    }
};
farm f[maxn];
int main()
{
#ifdef USACO    
    ofstream fout ("milk.out");
    ifstream fin ("milk.in");
#endif    
    ///变量定义
    while (cin >> N >> M)
    {
        for (int i = 0; i < M; i++)
        {
            cin >> f[i].price >> f[i].total;
        }
        sort(f, f + M);

        int remainsNeed = N, cost = 0;
        for (int i = 0; i < M; i++)
        {
            if (f[i].total > remainsNeed)
            {
                cost += f[i].price * remainsNeed;
                break;
            }
            cost += f[i].price * f[i].total;
            remainsNeed -= f[i].total;
        }
        cout << cost << endl;
    }

    ///结束
    return 0;
}