C++ priority_queue使用方法

以leetcode 1081题为例，https://leetcode.cn/problems/number-of-orders-in-the-backlog/

class Solution {
public:
    int getNumberOfBacklogOrders(vector<vector<int>>& orders) {
        // 新增sell订单，找buy的大于sell的，找buy最大值
        // 新增buy订单，找sell的小于buy的，找sell最小值
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> sell; // [price, amount]
        priority_queue<pair<int, int>, vector<pair<int, int>>, less<pair<int, int>>> buy; 
        for (auto od : orders) {
            if (od[2] == 0) { // buy
                while (!sell.empty() && od[1] > 0 && sell.top().first <= od[0]) {
                    if (od[1] >= sell.top().second) {
                        od[1] -= sell.top().second;
                        sell.pop();
                    } else {
                        int price = sell.top().first;
                        int amount = sell.top().second-od[1];
                        sell.pop();
                        sell.push({price, amount});
                        od[1] = 0;
                    }
                }
                if (od[1] != 0) {
                    buy.push({od[0], od[1]});
                }
                
            } else {  // sell
                while (!buy.empty() && od[1] > 0 && buy.top().first >= od[0]) {
                    if (od[1] >= buy.top().second) {
                        od[1] -= buy.top().second;
                        buy.pop();
                    } else {
                        int price = buy.top().first;
                        int amount = buy.top().second-od[1];
                        buy.pop();
                        buy.push({price, amount});
                        od[1] = 0;
                    }
                }
                if (od[1] != 0) {
                    sell.push({od[0], od[1]});
                }
            }
        }
        int MOD = 1000000007;
        int total = 0;
        while (!buy.empty()) {
            total = (total + buy.top().second) % MOD;
            buy.pop();
        }
        while (!sell.empty()) {
            total = (total + sell.top().second) % MOD;
            sell.pop();
        }
        return total;
    }
};