leetcode 5233 Maximum Profit in Job Scheduling

0/1区间dp。取一些不相交的区间。使得收益最大。对于一个区间[a, b], 其收益为v。对于b之后所有与[a,b]不相交的区间来说，v肯定能选，用lower_bound找出第一个大于等于b的区间的收益x，则选取该区间的收益为x+v。但是[a,b]不一定选，因为如果有区间和它相交，则肯定会被lower_bound跳过，所以x+v就要和当前的最大值mx比较。保存最大值。

class Solution {
public:int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {map<int, int> times; // times[i]表示到i结束的任务的最大收益for(int i : startTime)times[i] = 0;unordered_map<int, vector<pair<int, int>>> jobs;for(int i = 0; i < startTime.size(); i++)jobs[startTime[i]].push_back({endTime[i], profit[i]});int mx = 0;for(auto it = rbegin(times); it != rend(times); it++) // 根据结束时间从后往前算{for(auto job : jobs[it->first]){auto it = times.lower_bound(job.first); // 大于等于job结束时间能获得的最大收益mx = max(mx, job.second + (it == end(times) ? 0 : it->second));}it->second = mx;}return mx;}
};

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