Team Queue |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 60 Accepted Submission(s): 36 |
| |
|
Problem Description
Queues and Priority Queues are data structures which are known to
most computer scientists. The Team Queue, however, is not so well
known, though it occurs often in everyday life. At lunch time the queue
in front of the Mensa is a team queue, for example.
In a team queue each element belongs to a team. If an element enters the queue, it first searches the queue from head to tail to check if some of its teammates (elements of the same team) are already in the queue. If yes, it enters the queue right behind them. If not, it enters the queue at the tail and becomes the new last element (bad luck). Dequeuing is done like in normal queues: elements are processed from head to tail in the order they appear in the team queue. Your task is to write a program that simulates such a team queue. |
|
Input
The input will contain one or more test cases. Each test case
begins with the number of teams t (1<=t<=1000). Then t team
descriptions follow, each one consisting of the number of elements
belonging to the team and the elements themselves. Elements are integers
in the range 0 - 999999. A team may consist of up to 1000 elements.
Finally, a list of commands follows. There are three different kinds of commands: ENQUEUE x - enter element x into the team queue DEQUEUE - process the first element and remove it from the queue STOP - end of test case The input will be terminated by a value of 0 for t. |
|
Output
For each test case, first print a line saying "Scenario #k",
where k is the number of the test case. Then, for each DEQUEUE command,
print the element which is dequeued on a single line. Print a blank line
after each test case, even after the last one.
|
|
Sample Input
2 3 101 102 103 3 201 202 203 ENQUEUE 101 ENQUEUE 201 ENQUEUE 102 ENQUEUE 202 ENQUEUE 103 ENQUEUE 203 DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 2 5 259001 259002 259003 259004 259005 6 260001 260002 260003 260004 260005 260006 ENQUEUE 259001 ENQUEUE 260001 ENQUEUE 259002 ENQUEUE 259003 ENQUEUE 259004 ENQUEUE 259005 DEQUEUE DEQUEUE ENQUEUE 260002 ENQUEUE 260003 DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 0 |
|
Sample Output
Scenario #1 101 102 103 201 202 203 Scenario #2 259001 259002 259003 259004 259005 260001 |
很恶心的一道题，要hash
1 #include <cmath> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <string> 6 #include <cstdlib> 7 #include <queue> 8 using namespace std; 9 10 const int maxn=1000210; 11 struct qq 12 { 13 int x,t,num; 14 friend bool operator < (qq a,qq b) 15 { 16 if (a.num==b.num) 17 return a.t>b.t; 18 return a.num>b.num; 19 } 20 }ya; //num 表示在队列里它是属于第几个梯队的，t表示加入的时间，x表示这个数 21 priority_queue<qq> q; 22 int cnt,n,x,step; 23 int team[maxn],num[maxn],tot[maxn]; 24 char s[20]; 25 26 void close() 27 { 28 exit(0); 29 } 30 31 32 void init() 33 { 34 int cas=0; 35 while (scanf("%d",&n)!=EOF) 36 { 37 if (cas!=0) 38 printf(" "); 39 if (n==0) break; 40 while (!q.empty()) q.pop(); 41 cas++; 42 printf("Scenario #%d ",cas); 43 memset(num,-1,sizeof(num)); 44 for (int i=1;i<=n;i++) 45 { 46 scanf("%d",&cnt); 47 for (int j=1;j<=cnt;j++) 48 { 49 scanf("%d",&x); 50 team[x]=i; 51 } 52 } 53 cnt=0; 54 step=0; 55 memset(tot,0,sizeof(tot)); 56 while (scanf("%s",s)!=EOF) 57 { 58 step++; 59 if (s[0]=='S') break; 60 if (s[0]=='E') 61 { 62 scanf("%d",&x); 63 if (num[team[x]]!=-1 && tot[team[x]]!=0) //already in it 64 { 65 ya.num=num[team[x]]; 66 tot[team[x]]++; 67 } 68 else 69 { 70 cnt++; 71 ya.num=cnt; 72 num[team[x]]=cnt; 73 tot[team[x]]++; 74 } 75 ya.x=x; 76 ya.t=step; 77 q.push(ya); 78 } 79 else 80 { 81 ya=q.top(); 82 tot[team[ya.x]]--; 83 q.pop(); 84 printf("%d ",ya.x); 85 } 86 } 87 } 88 } 89 90 int main () 91 { 92 init(); 93 close(); 94 return 0; 95 }