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;
}