PAT 1026 Table Tennis[比較難]
1026 Table Tennis (30)（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 - 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
2Sample 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我的代碼：
#include <stdio.h> #include<iostream> #include <algorithm> using namespace std; struct Play{ int arrive,serve,ing,ed,vip; int waiting; Play(){ serve=-1;//表示沒被服務。 } }play[10000]; struct Table{ int vip,people; int ing;//是否正在使用 Table(){ vip=0;ing=-1;people=0; } }table[101]; bool cmp(Play& a,Play& b){ return a.arrive<b.arrive;//從小到大。 } bool cmp2(Play& a,Play& b){ return a.serve<b.serve; } int main() { int n,m,vm,vmno; scanf("%d",&n); int h,miu,s,bg=8*3600; for(int i=0;i<n;i++){ scanf("%d:%d:%d",&h,&miu,&s); play[i].arrive=h*3600+miu*60+s-bg; scanf("%d %d",&play[i].ing,&play[i].vip); play[i].ing*=60; if(play[i].ing>7200) play[i].ing=7200; } sort(play,play+n,cmp); scanf("%d%d",&m,&vm); for(int i=1;i<=vm;i++){ scanf("%d",&vmno); table[vmno].vip=1;//是vip桌子。 } // for(int i=0;i<n;i++){ // printf("%02d:%02d:%02d ",play[i].arrive/3600+8,play[i].arrive%3600/60,play[i].arrive%60); // printf("%d %d\n",play[i].ing,play[i].vip); // //printf("%02d:%02d:%02d ",play[i].serve/3600+8,play[i].serve%3600/60,play[i].serve%60); // } //一定要註意要服務一個人的時候，需要保證那個人已經到了。。。 //註意，題目要求，如果當前隊列中沒有VIP選手，那麽正常選手可以使用VIP桌子。 int ct=0; for(int tm=0;tm<46800;tm++){ //判斷當前是否有結束的 //不能用來計數，因為有可能serve到。 if(ct==n-1)break;//人已經分配完了。 for(int i=1;i<=m;i++){ if(table[i].ing!=-1&&play[table[i].ing].ed==tm){//判斷當前服務是否結束 table[i].ing=-1; } } int bg=0,j,k;//記錄開始循環的地方。 for(int i=1;i<=m;i++){ j=bg; if(table[i].ing==-1&&table[i].vip==1){//分配空閑vip桌子,分配vip用戶。 for(j=bg;j<n;j++){ if(play[j].arrive<=tm&&play[j].vip==1&&play[j].serve==-1){//當前選手是vip,且沒被服務 table[i].ing=j; table[i].people++; play[j].serve=tm; play[j].ed=tm+play[j].ing;ct++; // printf("%d %d %d %d vip\n",i,j,tm,tm-play[j].arrive); break;//跳出內層循環 } } bg=j; } } bg=0; for(int i=1;i<=m;i++){ k=bg; if(table[i].ing==-1){//分配桌子與群眾 for(k=bg;k<n;k++){ if(play[k].arrive<=tm&&play[k].serve==-1){ table[i].ing=k; table[i].people++; play[k].serve=tm; play[k].ed=tm+play[k].ing;ct++; // printf("%d %d %d %d\n",i,j,tm,tm-play[j].arrive); break; } } bg=k;//加了這個之後就可以了，不用每次都從頭開始遍歷。有一個標記。 } } } ct=0; sort(play,play+n,cmp2); for(int i=0;i<n;i++){ if(play[i].serve!=-1){ printf("%02d:%02d:%02d ",play[i].arrive/3600+8,play[i].arrive%3600/60,play[i].arrive%60); printf("%02d:%02d:%02d ",play[i].serve/3600+8,play[i].serve%3600/60,play[i].serve%60); if((play[i].serve-play[i].arrive)%60>=30) ct=1; else ct=0; printf("%d\n",(play[i].serve-play[i].arrive)%3600/60+ct); } } printf("%d",table[1].people); for(int i=2;i<=m;i++){ printf(" %d",table[i].people); } return 0; }
//！！！在牛客網上AC 了，但是在pat上就只有16分，只通過了兩個點，其他的都是答案錯誤，真是哭唧唧，完全不知道該怎麽改了。！！先放這。明天復習一下。絕望了。
其中第一次提交通過80%，有如下沒通過：
測試用例:
4
08:00:00 60 0
08:10:00 50 0
08:50:00 30 0
09:00:00 30 1
2 1
2
對應輸出應該為:
08:00:00 08:00:00 0
08:10:00 08:10:00 0
09:00:00 09:00:00 0
08:50:00 09:00:00 10
2 2
你的輸出為:
08:00:00 08:00:00 0
08:10:00 08:10:00 0
08:50:00 09:00:00 10
改了cmp2函數之後，提交之後報：您的程序未能在規定時間內運行結束，請檢查是否循環有錯或算法復雜度過大。
看來我這麽循環著做，不太對啊。復雜度太高了，一直循環。
PAT 1026 Table Tennis[比較難]