1026. Table Tennis (30)
時間限制
400 ms
記憶體限制
65536 kB
代碼長度限制
16000 B
判題程式
Standard
作者
CHEN, Yue
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 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 又是模拟排隊的問題，此題坑點很多，我在代碼中打了注釋的地方都是需要注意的。
不得不說，此題的資料應該是水了，對于到達時有多個桌子空出的情況我一開始選擇的是最小的，竟然可以對。
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<iostream>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int maxn = 1e5 + 10;
int n, f[maxn], t[maxn], x, y, z, m, u, vis[maxn], cnt[maxn];
struct point
{
int x, y, f;
bool operator<(const point &a)const
{
return x < a.x;
}
}a[maxn];
void putout(int x)
{
printf("%02d:%02d:%02d ", x / 3600, x / 60 % 60, x % 60);
}
void putmin(int x)
{
printf("%d\n", x / 60 + (x % 60 < 30 ? 0 : 1));//滿30秒進1分
}
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d:%d:%d", &x, &y, &z);
a[i].x = x * 3600 + y * 60 + z;
scanf("%d%d", &x, &a[i].f);
a[i].y = min(x * 60, 7200);//最多不超過2小時
}
sort(a, a + n);
scanf("%d%d", &m, &u);
while (u--) scanf("%d", &x), f[x] = 1;
for (int i = 1; i <= m; i++) t[i] = 8 * 3600;
for (int i = 0; i < n; i++)
if (!vis[i])
{
int now = 1;
for (int j = 1; j <= m; j++) if (t[now] > t[j]) now = j;//找出最早空出的桌子
if (max(t[now], a[i].x) >= 21 * 3600) break;//超過21點跳出
if (t[now] <= a[i].x)//這意味着現在不用排隊
{
int vip = -1;//如果這個人是vip那麼會選擇能用的vip的桌子中最小的
for (int j = m; j >= 1; j--)
{
if (t[j] <= a[i].x) now = j;
if (t[j] <= a[i].x&&f[j]) vip = j;
}
if (a[i].f&&vip != -1) now = vip; //如果這個人是vip且有vip的桌子能用
cnt[now]++;
putout(a[i].x); putout(a[i].x);
printf("0\n"); t[now] = a[i].x + a[i].y;
}
else//這意味着現在有人在排隊
{
/*這部分有個問題
如果有人在排隊，并且同時空出了一張非vip桌子和一張vip桌子，
如果标号小的不是vip，标号大的是vip，那麼先處理哪個呢，題目沒有明确的指出
經過我的測試，也沒有這樣的資料，是以可以忽略。
int vip = -1; //如果有vip桌子同時也可用
for (int j = m; j >= 1; j--)
{
if (t[j] == t[now] && f[j]) vip = j;
}
if (vip != -1)//先選擇處理vip桌子
{
int flag = 0;
for (int j = i; j < n&&a[j].x <= t[vip]; j++)
{
if (a[j].f&&!vis[j])
{
vis[j] = 1;
putout(a[j].x); putout(t[vip]);
putmin(t[vip] - a[j].x);
t[vip] = t[vip] + a[j].y;
flag = 1; break;
}
}
if (flag) { cnt[vip]++; i--; continue; }
}
*/
cnt[now]++;
if (f[now]) //如果是vip的桌子
{
int flag = 0;
for (int j = i; j < n&&a[j].x <= t[now]; j++)//找到是否有vip在等待
{
if (a[j].f&&!vis[j])
{
vis[j] = 1;
putout(a[j].x); putout(t[now]);
putmin(t[now] - a[j].x);
t[now] = t[now] + a[j].y;
flag = 1; break;
}
}
if (flag) i--;
else//如果沒有，那麼第一個人用了
{
putout(a[i].x); putout(t[now]);
putmin(t[now] - a[i].x);
t[now] = t[now] + a[i].y;
}
}
else//非vip桌子，直接用
{
putout(a[i].x); putout(t[now]);
putmin(t[now] - a[i].x);
t[now] = t[now] + a[i].y;
}
}
}
for (int i = 1; i <= m; i++) printf("%d%s", cnt[i], i == m ? "\n" : " ");
return 0;
}