[Leetcode] 502. IPO 解题报告

题目：

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 most k distinct
projects before the IPO. Help LeetCode design the best way to maximize its total capital after finishing at most k distinct projects.

You are given several projects. For each project i, it has a pure profit Pi and
a minimum capital of Ci is needed to start the corresponding project. Initially, you have W 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: 4

Explanation: 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.
思路：

一道贪心算法的题目：每次我们都选在当前资金允许的条件下，可以获利最大的项目。所以我们首先将项目按照所需资金进行排序（需要资金越小的项目越放在前面）。然后建立一个优先队列（priority_queue）来存储目前可以做的项目，每当需要选择一个项目的时候，就从项目队列中选出在目前资金状况下可以做的项目，然后加入到优先队列中；最后在优先队列中选出获利最大的一个项目来做（位于队列首部），并更新当前资金。直到k个项目完成为止。

算法的时间复杂度是O(nlogn)，空间复杂度是O(n)。

代码：

class Solution {
public:
int findMaximizedCapital(int k, int W, vector<int>& Profits, vector<int>& Capital) {
vector<pair<int, int>> capital_profits; // pair of (Captical, Profits)
for (int i = 0; i < Capital.size(); ++i) {
capital_profits.push_back(make_pair(Capital[i], Profits[i]));
}
sort(capital_profits.begin(), capital_profits.end(), PairComp);
int index = 0;
priority_queue<pair<int, int>> projects;
for (int i = 0; i < k; ++i) {
for (; index < capital_profits.size(); ++index) { // add the projects whose capital are not larger than W
if (W >= capital_profits[index].first) {
projects.push(make_pair(capital_profits[index].second, capital_profits[index].first));
}
else {
break;
}
}
if (projects.size() > 0) { // do the project that has the maximal profits
pair<int, int> profit_capital = projects.top();
projects.pop();
W += profit_capital.first;
}
}
return W;
}
private:
struct PairCompare {
bool operator() (const pair<int, int> &a, const pair<int, int> &b) const {
return a.first != b.first ? a.first < b.first : a.second < b.second;
}
} PairComp;
};