PAT甲級 1026 Table Tennis（30 分）模擬，排序 ----------
1026 Table Tennis（30 分）
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
K (≤100) - the number of tables, and M (< K) - the number of VIP tables. The last line contains M table 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題目大意：k張桌子，球員到達後總是選擇編號最小的桌子。如果訓練時間超過2h會被壓縮成2h，如果到達時候沒有球桌空閒就變成佇列等待。
k張桌子中m張是vip桌，如果vip桌子有空閒，而且佇列裡面有vip成員，那麼等待佇列中的第一個vip球員會到最小的vip球桌訓練。如果vip桌子空閒但是沒有vip來，那麼就分配給普通的人。如果沒有vip球桌空閒，那麼vip球員就當作普通人處理。
給出每個球員的到達時間、要玩多久、是不是vip（是為1不是為0）。給出球桌數和所有vip球桌的編號，QQ所有在關門前得到訓練的球員的到達時間、訓練開始時間、等待時長（取整數，四捨五入），營業時間為8點到21點。如果再21點後還沒有開始玩的人，就不再玩，不需要輸出～
分析：建立兩個結構體，person 和 tablenode，分別建立他們的結構體陣列player和table。
因為給出的輸入是無序的所以要先排序。可以根據最早空閒的桌子是不是vip桌進行分類討論。
如果最早空閒的桌子index是vip桌，那麼尋找佇列裡面第一個vip球員。根據他的到達時間繼續分類討論。
如果最早空閒的桌子index不是vip桌，那麼看隊首的球員是不是vip球員。如果是普通人，就直接把球桌分配給他，如果是vip，那麼需要找到最早空閒的vip桌子的號vipindex，根據vip球員的到達時間和vipindex桌子的空閒時間繼續分類討論。
分配的時候標記table的num++，統計該table服務的人數。
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
struct person {
int arrive, start, time;
bool vip;
}tempperson;
struct tablenode {
int end = 8 * 3600, num;
bool vip;
};
bool cmp1(person a, person b) {
return a.arrive < b.arrive;
}
bool cmp2(person a, person b) {
return a.start < b.start;
}
vector<person> player;
vector<tablenode> table;
void alloctable(int personid, int tableid) {
if(player[personid].arrive <= table[tableid].end)
player[personid].start = table[tableid].end;
else
player[personid].start = player[personid].arrive;
table[tableid].end = player[personid].start + player[personid].time;
table[tableid].num++;
}
int findnextvip(int vipid) {
vipid++;
while(vipid < player.size() && player[vipid].vip == false) {
vipid++;
}
return vipid;
}
int main() {
int n, k, m, viptable;
scanf("%d", &n);
for(int i = 0; i < n; i++) {
int h, m, s, temptime, flag;
scanf("%d:%d:%d %d %d", &h, &m, &s, &temptime, &flag);
tempperson.arrive = h * 3600 + m * 60 + s;
tempperson.start = 21 * 3600;
if(tempperson.arrive >= 21 * 3600)
continue;
tempperson.time = temptime <= 120 ? temptime * 60 : 7200;
tempperson.vip = ((flag == 1) ? true : false);
player.push_back(tempperson);
}
scanf("%d%d", &k, &m);
table.resize(k + 1);
for(int i = 0; i < m; i++) {
scanf("%d", &viptable);
table[viptable].vip = true;
}
sort(player.begin(), player.end(), cmp1);
int i = 0, vipid = -1;
vipid = findnextvip(vipid);
while(i < player.size()) {
int index = -1, minendtime = 999999999;
for(int j = 1; j <= k; j++) {
if(table[j].end < minendtime) {
minendtime = table[j].end;
index = j;
}
}
if(table[index].end >= 21 * 3600)
break;
if(player[i].vip == true && i < vipid) {
i++;
continue;
}
if(table[index].vip == true) {
if(player[i].vip == true) {
alloctable(i, index);
if(vipid == i)
vipid = findnextvip(vipid);
i++;
} else {
if(vipid < player.size() && player[vipid].arrive <= table[index].end) {
alloctable(vipid, index);
vipid = findnextvip(vipid);
} else {
alloctable(i, index);
i++;
}
}
} else {
if(player[i].vip == false) {
alloctable(i, index);
i++;
} else {
int vipindex = -1, minvipendtime = 999999999;
for(int j = 1; j <= k; j++) {
if(table[j].vip == true && table[j].end < minvipendtime) {
minvipendtime = table[j].end;
vipindex = j;
}
}
if(vipindex != -1 && player[i].arrive >= table[vipindex].end) {
alloctable(i, vipindex);
if(vipid == i)
vipid = findnextvip(vipid);
i++;
} else {
alloctable(i, index);
if(vipid == i)
vipid = findnextvip(vipid);
i++;
}
}
}
}
sort(player.begin(), player.end(), cmp2);
for(i = 0; i < player.size() && player[i].start < 21 * 3600; i++) {
printf("%02d:%02d:%02d ", player[i].arrive / 3600, player[i].arrive % 3600 / 60, player[i].arrive % 60);
printf("%02d:%02d:%02d ", player[i].start / 3600, player[i].start % 3600 / 60, player[i].start % 60);
printf("%.0f\n", round((player[i].start - player[i].arrive) / 60.0));
}
for(int i = 1; i <= k; i++) {
if(i != 1)
printf(" ");
printf("%d", table[i].num);
}
return 0;
}