UVA 12100 Printer Queue

最新推荐文章于 2021-01-22 21:42:27 发布

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本文介绍了一种解决打印机队列问题的方法，通过使用优先队列和重载关系运算符来确定打印作业完成的时间。示例输入输出展示了算法的有效性。

12100 Printer Queue

Input

One line with a positive integer : the number of test cases(at most 100). Then for each test case:

• One line with two integers n and m, where n is the number of jobs in the queue (1≤n≤100)

And m is the position of your job (0≤m≤n−1). The ﬁrst position in the queue is number 0,

the second is number 1,and soon.

• One line with n integers in the range 1 to 9, giving the priorities of the jobs in the queue. The

ﬁrst integer gives the priority of the ﬁrst job, the second integer the priority of the second job,

and soon.

Output

For each test case,print one line with a single integer;the number of minutes until your job is completely

printed,as suming that no additional print jobs will arrive.

Sample Input

1 0

4 2

1 2 3 4

6 0

1 1 9 1 1 1

Sample Output

用到了关系运算符的重载

关系运算符的定义

关系运算符有==,!=,<,>,<=,>=。

bool operator == (const A& ); 
bool operator != (const A& );
bool operator < (const A& );
bool operator <= (const A& );
bool operator > (const A& );
bool operator >= (const A& );

还有就是优先队列的使用，因为是结构体所以要用到运算符重载。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
struct node
{
    int num,level;
    bool operator<(const node &a)const//关系运算符<的重载
    {
        return level<a.level;
    }
};
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        queue<node>q;
        priority_queue<node>qq;
        struct node a;
        int n,m;
        cin>>n>>m;
        int i;
        for(i=0;i<=n-1;i++)
        {
            scanf("%d",&a.level);
            a.num=i;
            q.push(a);
            qq.push(a);
        }
        int ans=0;
        while(1)
        {
            if(qq.top().level==q.front().level)
            {
                ans++;
                a=q.front();
                qq.pop();
                q.pop();
                if(a.num==m) break;
            }
            else
            {
                a=q.front();
                q.pop();
                q.push(a);
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}