部分(((φ(◎ロ◎;)φ)))

• vector使用sort排序：需要头文件#include<algorithm>，
• 喵

sort(vec.begin(),vec.end());(默认是按升序排列,即从小到大).

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue> // n^复杂度计算10000次，时间限制0.4s，为了不超时，球员要用堆，球台可以不用100，100————服务10000,遍历100次复杂度100*10000
using namespace std;

const int N = 10010, int M = 110;

struct Player
{
    int arr_time;
    int play_time;
    int begin_time;
    int wait_time;

    bool operator< (const Player& t) const //⭐⭐sort排序，堆用大于号
    {
        if (begin_time != t.begin_time) begin_time < t.begin_time;
        return arr_time < t.arr_time;
    }

    bool operator> (const Player& t) const //堆排序用大于号，优先队列中比较大小
    {
        return arr_time > t.arr_time;
    }
}players[N];

struct Table
{
    int nums_playerstable;
    int end_time;

    bool operator> (const Table& t) const //堆排序用 > ⭐
    {
        if (end_time != t.end_time) return end_time > t.end_time;
        return id > t.id;
    }
}

priority_queue<Player, vector<Player>, greater<Player>> normal_player;
priority_queue<Player, vector<Player>, greater<Player>> vip_player;

int mums_player, vip;

int mian()
{
    cin >> nums_player;

    int hour, minute, second, p_t, vip;
    for (int i = 0; i < nums_player; i ++ )
    {
        scanf("%d:%d:%d %d %d", &hour, &minute, &second, &p_t, &vip );

        int day_time = hour * 3600 + minute * 60 + second;

        if(vip) vip_player.push_back({day_time, p_t});
        else normal_player.push_back({day_time, p_t});
    }


}