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UVA-540 Team Queue
阅读量:7248 次
发布时间:2019-06-29

本文共 2642 字,大约阅读时间需要 8 分钟。

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 file 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.
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.

代码:

#include<iostream>

#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
using namespace std;
const int N = 1000 + 5;
void solve_question(int n){
    map<int,int> group;
    int m,x;
    for(int i=0;i<n;i++){
        scanf("%d",&m);
        for(int j=0;j<m;j++) {scanf("%d",&x); group[x] = i;}
    }
    queue<int> q,q2[N];
    char ch[10];
    while(true){
        scanf("%s",ch);
        if(ch[0] == 'S') break;
        if(ch[0] == 'E'){
            scanf("%d",&x);
            int t = group[x];
            if(q2[t].empty()) q.push(t);
            q2[t].push(x);
        }else{
            int t = q.front();
            printf("%d\n",q2[t].front());q2[t].pop();
            if(q2[t].empty()) q.pop();
        }
    }
}
int main(){
    int n,Case=0;
    while(scanf("%d",&n)==1 && n){
        printf("Scenario #%d\n",++Case);
        solve_question(n);
        printf("\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/Pretty9/p/7347674.html

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