502. IPO

Suppose LeetCode will start its ipO soon. In order to sell a good PRice of its shares to Venture Capital, LeetCode would like to work on some projects to increase its capital before the IPO. Since it has limited resources, it can only finish at mostk distinct projects before the IPO. Help LeetCode design the best way to maximize its total capital after finishing at mostk distinct projects.

You are given several projects. For each project i, it has a pure profitP_i and a minimum capital of C_i is needed to start the corresponding project. Initially, you haveW capital. When you finish a project, you will obtain its pure profit and the profit will be added to your total capital.

To sum up, pick a list of at most k distinct projects from given projects to maximize your final capital, and output your final maximized capital.

Example 1:

Input: k=2, W=0, Profits=[1,2,3], Capital=[0,1,1].Output: 4Explanation: Since your initial capital is 0, you can only start the project indexed 0.             After finishing it you will obtain profit 1 and your capital becomes 1.             With capital 1, you can either start the project indexed 1 or the project indexed 2.             Since you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital.             Therefore, output the final maximized capital, which is 0 + 1 + 3 = 4.
Note:
You may assume all numbers in the input are non-negative integers.The length of Profits array and Capital array will not exceed 50,000.The answer is guaranteed to fit in a 32-bit signed integer.
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初始资金为W，最多进行k次项目，后面给出n个项目的收益和需要的资金，问最大的收益。用贪心算法。对于每一次项目，都选最优的做，即选择需要资金小于或等于当前资金的项目中收益最大的。为了减少搜索的时间，用两个优先队列，其中pq1按需要资金从小到大排序，一开始存放全部的项目（项目用（需要资金，收益）来表示）；另一个pq2按收益 从大到小排序，存放当前能做的项目。对于某一次选择，将pq1中需要资金小于等于当前资金的项目pop出来push进pq2，然后选择pq2的top项目（也要pop出来），该项目就是当前最佳的项目。
代码：
class Solution{public:	int findMaximizedCapital(int k, int W, vector<int>& Profits, vector<int>& Capital)	{		auto cmp1 = [](const pair<int, int>& lhs, const pair<int, int>& rhs) -> bool { return lhs.first > rhs.first; };		priority_queue<pair<int, int>, vector<pair<int, int> >, function<bool (const pair<int, int>&, const pair<int, int>&)> > pq1(cmp1);		auto cmp2 = [](const pair<int, int>& lhs, const pair<int, int>& rhs) -> bool { return lhs.second < rhs.second; };		priority_queue<pair<int, int>, vector<pair<int, int> >, function<bool (const pair<int, int>&, const pair<int, int>&)> > pq2(cmp2);		for(int i = 0; i < Profits.size(); ++i)		{			pq1.push(pair<int, int>(Capital[i], Profits[i]));		}		int res = W;		while(--k >= 0)		{			while(!pq1.empty() && pq1.top().first <= res)			{				pq2.push(pq1.top());				pq1.pop();			}			if(pq2.empty()) break;			res += pq2.top().second;			pq2.pop();		}		return res;	}};