leetcode1882 使用服务器处理任务

思路：

维护两个堆分别存储空闲的服务器和正在使用的服务器，遍历任务分别进行处理。均摊时间复杂度n*log(m)（n是任务数量，m是服务器数量）。

实现：

class Solution
{
public:
    vector<int> assignTasks(vector<int>& servers, vector<int>& tasks)
    {
        int n = servers.size(), m = tasks.size();
        using pii = pair<int, int>;
        using tup = tuple<int, int, int>;
        priority_queue<tup, vector<tup>, greater<tup>> t;
        priority_queue<pii, vector<pii>, greater<pii>> q;
        for (int i = 0; i < n; i++)
        {
            t.push({0, servers[i], i});
        }
        vector<int> res;
        for (int i = 0; i < m; i++)
        {
            while (!t.empty())
            {
                auto tmp = t.top();
                int time = get<0>(tmp), weight = get<1>(tmp), server_id = get<2>(tmp);
                if (time > i) break;
                t.pop();
                q.push({weight, server_id});
            }
            if (q.empty())
            {
                auto tmp = t.top(); t.pop();
                int time = get<0>(tmp), weight = get<1>(tmp), server_id = get<2>(tmp);
                res.push_back(server_id);
                t.push({time + tasks[i], weight, server_id});
            }
            else
            {
                auto tmp = q.top(); q.pop();
                int server_id = tmp.second; 
                res.push_back(server_id);
                t.push({i + tasks[i], servers[server_id], server_id});
            }
        }
        return res;
    }
};