PAT A1026 Table Tennis （30 分）——队列

A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For any pair of players, if there are some tables open when they arrive, they will be assigned to the available table with the smallest number. If all the tables are occupied, they will have to wait in a queue. It is assumed that every pair of players can play for at most 2 hours.

Your job is to count for everyone in queue their waiting time, and for each table the number of players it has served for the day.

One thing that makes this procedure a bit complicated is that the club reserves some tables for their VIP members. When a VIP table is open, the first VIP pair in the queue will have the priviledge to take it. However, if there is no VIP in the queue, the next pair of players can take it. On the other hand, if when it is the turn of a VIP pair, yet no VIP table is available, they can be assigned as any ordinary players.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (≤10000) - the total number of pairs of players. Then N lines follow, each contains 2 times and a VIP tag: HH:MM:SS - the arriving time, P - the playing time in minutes of a pair of players, and tag - which is 1 if they hold a VIP card, or 0 if not. It is guaranteed that the arriving time is between 08:00:00 and 21:00:00 while the club is open. It is assumed that no two customers arrives at the same time. Following the players' info, there are 2 positive integers: K (≤100) - the number of tables, and M (< K) - the number of VIP tables. The last line contains Mtable numbers.

Output Specification:

For each test case, first print the arriving time, serving time and the waiting time for each pair of players in the format shown by the sample. Then print in a line the number of players served by each table. Notice that the output must be listed in chronological order of the serving time. The waiting time must be rounded up to an integer minute(s). If one cannot get a table before the closing time, their information must NOT be printed.

Sample Input:

9
20:52:00 10 0
08:00:00 20 0
08:02:00 30 0
20:51:00 10 0
08:10:00 5 0
08:12:00 10 1
20:50:00 10 0
08:01:30 15 1
20:53:00 10 1
3 1
2

Sample Output:

08:00:00 08:00:00 0
08:01:30 08:01:30 0
08:02:00 08:02:00 0
08:12:00 08:16:30 5
08:10:00 08:20:00 10
20:50:00 20:50:00 0
20:51:00 20:51:00 0
20:52:00 20:52:00 0
3 3 2

#include <stdio.h>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn = ;
struct window {
    int isvip=;
    int count=;
    int serve=*;
}win[maxn];
struct pairs {
    int h;
    int m;
    int s;
    int arrive;
    int time;
    int vip;
    int wait;
    int serve;
};
int n, k, m;
vector<pairs> v;
queue<pairs> q;
bool cmp(pairs p1, pairs p2) {
    return p1.arrive < p2.arrive;
}
int main() {
    scanf("%d", &n);
    int flag = ;
    for (int i = ; i < n; i++) {
        pairs p;
        scanf("%d:%d:%d %d %d", &p.h, &p.m, &p.s, &p.time, &p.vip);
        p.arrive = p.h *  + p.m *  + p.s;
        if (p.time > )p.time = ;
        p.time *= ;
        v.push_back(p);
        if (p.vip == )flag++;
    }
    scanf("%d %d", &k, &m);
    for (int i = ; i < m; i++) {
        int j;
        scanf("%d", &j);
        win[j].isvip = ;
    }
    sort(v.begin(), v.end(), cmp);
    for (int i = ; i < n; i++) {
        if (v[i].arrive >=  * )break;
        int min =  * ;
        int min_j = ;
        for (int j = ; j <= k; j++) {
            if (win[j].serve < min) {
                min = win[j].serve;
                min_j = j;
            }
        }
        if (min_j != ) {
            if (v[i].arrive >= min) {
                if (v[i].vip == ) {
                    int vip_j = ;
                    for (int j = ; j <= k; j++) {
                        if (win[j].serve <= v[i].arrive && win[j].isvip == ) {
                            vip_j = j;
                            break;
                        }
                    }
                    if (vip_j != ) {
                        win[vip_j].serve = v[i].arrive;
                        v[i].wait = ;
                        v[i].serve = win[vip_j].serve;
                        win[vip_j].serve += v[i].time;
                        win[vip_j].count++;
                        if (v[i].serve <  * )q.push(v[i]);
                    }
                    else {
                        int vip_j = ;
                        for (int j = ; j <= k; j++) {
                            if (win[j].serve <= v[i].arrive) {
                                vip_j = j;
                                break;
                            }
                        }
                        if (vip_j != ) {
                            win[vip_j].serve = v[i].arrive;
                            v[i].wait = ;
                            v[i].serve = win[vip_j].serve;
                            win[vip_j].serve += v[i].time;
                            win[vip_j].count++;
                            if (v[i].serve <  * )q.push(v[i]);
                        }
                    }
                }
                else {
                    int vip_j = ;
                    for (int j = ; j <= k; j++) {
                        if (win[j].serve <= v[i].arrive) {
                            vip_j = j;
                            break;
                        }
                    }
                    if (vip_j != ) {
                        win[vip_j].serve = v[i].arrive;
                        v[i].wait = ;
                        v[i].serve = win[vip_j].serve;
                        win[vip_j].serve += v[i].time;
                        win[vip_j].count++;
                        if (v[i].serve <  * )q.push(v[i]);
                    }
                }
            }
            else {
                if (win[min_j].isvip == ) {
                    v[i].wait = (min - v[i].arrive) /  + ((min - v[i].arrive) %  >=  ?  : );
                    v[i].serve = win[min_j].serve;
                    win[min_j].serve += v[i].time;
                    win[min_j].count++;
                    if (v[i].serve <  * )q.push(v[i]);
                }
                else {
                    int tmp_j = i;
                    for (tmp_j; tmp_j < n; tmp_j++) {
                        if (v[tmp_j].vip ==  && v[tmp_j].arrive <= min) {
                            pairs tmp_p2 = v[tmp_j];
                            v.erase(v.begin() + tmp_j);
                            v.insert(v.begin() + i, tmp_p2);
                            break;
                        }
                    }
                    v[i].wait = (min - v[i].arrive) /  + ((min - v[i].arrive) %  >=  ?  : );
                    v[i].serve = win[min_j].serve;
                    win[min_j].serve += v[i].time;
                    win[min_j].count++;
                    if (v[i].serve <  * )q.push(v[i]);
                }
            }
        }
        else break;
    }
    while (!q.empty()) {
        pairs qp = q.front();
        q.pop();
        printf("%02d:%02d:%02d %02d:%02d:%02d %d\n", qp.h, qp.m, qp.s, qp.serve / , qp.serve %  / , qp.serve %  % , qp.wait);
    }
    for (int i = ; i <= k; i++) {
        printf("%d", win[i].count);
        if (i != k)printf(" ");
    }
    system("pause");
}

注意点：测试点3是有一个21点来的人，题目说保证到达时间在8点到21点之间，没说是小于，有刚好等于的，坑。

测试点4是有人想打超过2个小时，题目说了只能打最多2小时。

还有就是如果到的时候有空位，不是去最早服务的那个，而是要去桌号最小的满足服务时间那个桌

最后一点，一开始觉得题目很复杂，其实只要一点点写，写着写着就做出来了。我是先写出如果没有vip这种情况，再考虑有vip的时候要怎么修改。