[POJ2259] Team Queue
题目链接 :POJ2259;
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.
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:
The input will be terminated by a value of 0 for t.
Warning: A test case may contain up to 200000 (two hundred thousand) commands, so the implementation of the team queue should be efficient: both enqueing and dequeuing of an element should only take constant time.
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.
Warning: A test case may contain up to 200000 (two hundred thousand) commands, so the implementation of the team queue should be efficient: both enqueing and dequeuing of an element should only take constant time.
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
高考后第一次做题。
挺水的队列题, 维护每个团队现在在排队的人队列,再加一个每个团队顺序的队列(即que[0])。
dequeue的时候直接把第一个团队里的第一个人弹出来就行了。
水题练手。
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #include <map> using namespace std; #define reg register inline int read() { int res = 0;char ch = getchar();bool fu = 0; while(!isdigit(ch)) fu |= (ch == '-'), ch = getchar(); while(isdigit(ch)) res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar(); return fu ? -res : res; } int Time, n; queue<int> que[1005]; map<int, int> mp; int main() { while(1) { n = read(); if (!n) break; printf("Scenario #%d\n", ++Time); for (reg int i = 0 ; i <= 1000 ; i ++) while(!que[i].empty()) que[i].pop(); mp.clear(); for (reg int i = 1 ; i <= n ; i ++) { int num = read(); for (reg int j = 1 ; j <= num ; j ++) { mp[read()] = i; } } while(1) { string str; cin >> str; if (str[0] == 'S') break; else if (str[0] == 'E') { int x = read(), bel = mp[x]; if (que[bel].empty()) que[0].push(bel); que[bel].push(x); }else { int x = que[0].front(); printf("%d\n", que[x].front()); que[x].pop(); if (que[x].empty()) que[0].pop(); } } puts(""); } return 0; }