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.
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.
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
Scenario #1 101 102 103 201 202 203 Scenario #2 259001 259002 259003 259004 259005 260001
分析:
用到两个队列,一个是每个团队自己是一个队列,而每个团队整体之间又构成一个队列。
#include <iostream> #include <queue> #include <cstdio> #include <cstring> #include <map> #define maxn 1010 using namespace std; int main() { int kase = 0, t; while(scanf("%d", &t) == 1 && t) { printf("Scenario #%d\n", ++kase); queue<int> q; //表示由团队组成的队列 queue<int> team_q[maxn]; //team_q[i]表示编号为i的团队的队列 map<int, int> team; //team[i]表示在队列中编号为i的成员所在的团队编号 for(int i = 0; i < t; i++) { //之所以用map容器而不直接用数组来映射的原因是用int数组可能会导致内存开销过大 int n, x; //提交到oj上会产生runtime error scanf("%d", &n); while(n--) { scanf("%d", &x); //在团队中的编号为x的成员 team[x] = i; //将其所属团队的编号设置为i } } while(1) { char cmd[10]; scanf("%s", cmd); if(cmd[0] == 'S') break; else if(cmd[0] == 'E') { int x; scanf("%d", &x); int i = team[x]; //找到编号为x的队列成员所属的团队编号 if(team_q[i].empty()) q.push(i); team_q[i].push(x); } else if(cmd[0] == 'D') { int x = q.front(); //找到队列中最前列即将出列的成员所属团队的编号 printf("%d\n", team_q[x].front()); team_q[x].pop(); if(team_q[x].empty()) q.pop(); //如果出列的是该团队的最后一名成员,则在q中将整个团队pop() } } printf("\n"); } return 0; }