题目描述
Bessie and the herd love chocolate so Farmer John is buying them some.
The Bovine Chocolate Store features N (1 <= N <= 100,000) kinds of chocolate in essentially unlimited quantities. Each type i of chocolate has price P_i (1 <= P_i <= 10^18) per piece and there are C_i (1 <= C_i <= 10^18) cows that want that type of chocolate.
Farmer John has a budget of B (1 <= B <= 10^18) that he can spend on chocolates for the cows. What is the maximum number of cows that he can satisfy? All cows only want one type of chocolate, and will be satisfied only by that type.
Consider an example where FJ has 50 to spend on 5 different types of chocolate. A total of eleven cows have various chocolate preferences:
Chocolate_Type Per_Chocolate_Cost Cows_preferring_this_type 1 5 3
2 1 1
3 10 4
4 7 2
5 60 1
Obviously, FJ can't purchase chocolate type 5, since he doesn't have enough money. Even if it cost only 50, it's a counterproductive purchase since only one cow would be satisfied.
Looking at the chocolates start at the less expensive ones, he can * purchase 1 chocolate of type #2 for 1 x 1 leaving 50- 1=49, then * purchase 3 chocolate of type #1 for 3 x 5 leaving 49-15=34, then * purchase 2 chocolate of type #4 for 2 x 7 leaving 34-14=20, then * purchase 2 chocolate of type #3 for 2 x 10 leaving 20-20= 0.
He would thus satisfy 1 + 3 + 2 + 2 = 8 cows.
贝西和其他奶牛们都喜欢巧克力，所以约翰准备买一些送给她们。奶牛巧克力专卖店里
有N种巧克力，每种巧克力的数量都是无限多的。每头奶牛只喜欢一种巧克力，调查显示，
有Ci头奶牛喜欢第i种巧克力，这种巧克力的售价是P。
约翰手上有B元预算，怎样用这些钱让尽量多的奶牛高兴呢？
输入输出格式
输入格式：
* Line 1: Two space separated integers: N and B
* Lines 2..N+1: Line i contains two space separated integers defining chocolate type i: P_i and C_i
输出格式：
* Line 1: A single integer that is the maximum number of cows that Farmer John can satisfy
输入输出样例
输入样例#1
5 50 5 3 1 1 10 4 7 2 60 1
输出样例#1
8
思路
不一定要全部满足第ii种巧克力的奶牛，可以只满足一半。这是可以用贪心而不是用dp背包的理由。
尽量用价值最小的巧克力喂奶牛。这样会更省钱。
#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef struct
{
long long int like,value;
}node;
bool operator < (const node &a,const node &b)//排序规则
{
return a.like<b.like;
}
node a[100001];
long long int s,n,b;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
long long int i,j;
cin>>n>>b;
for(i=1;i<=n;i++)
{
cin>>a[i].like>>a[i].value;
}
sort(a+1,a+n+1);//价格从小往大排序
for(i=1;i<=n;i++)
{
if(a[i].like<=b)//剩下的钱是否能购买一个巧克力
{
long long int k=min(b/a[i].like,a[i].value);//不断买这种巧克力，能卖多少就买多少
b=b-k*a[i].like;//减掉花掉的钱
s=s+k;//加上满足的奶牛数
}
else//小小的优化，如果没钱就不必再搜索了
break;
}
cout<<s<<endl;
return 0;
}