Luo Valley P2278 [HNOI2003] operating system (analog, priority queue)

Portal

Problem-solving ideas

The first is the practice of violence :

From the beginning enumeration time, plus each time a judge has no new task arrives, then pick out the biggest priority value priority queue, perform a second, back in the queue.

Apparently TLE's!

Then think about optimization:

Where you can look for optimization, we found that only time! So every time we remove the head of the queue (task a) There are two cases - the first time the task is now time + a remaining time to complete the work required <next to arrive, this time the optimal strategy must be this is a complete piece of work; the second is to finish the job, do it as much as possible, the time required to subtract a surplus of time to do, and then back into the queue.

According to that when a new mission did not arrive, the best is always the first team, at that moment a new task arrives, not necessarily, because the new element may be higher than the team's first priority.

note

Weight carrying structure operator wording
Comparison can not be completed is not less than or less

AC Code

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cmath>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<queue>
 7 using namespace std;
 8 struct node{
 9     int num,t1,t2,d;
10     bool operator < (const node &x)const{
11         return d==x.d?t1>x.t1:d<x.d;
12     }
13 }x;
14 priority_queue<node> q;
15 int nowt;
16 int main()
17 {
18     while(scanf("%d",&x.num)!=EOF){
19         scanf("%d%d%d",&x.t1,&x.t2,&x.d);
20         if(q.empty()){
21             q.push(x);
22             nowt=x.t1;
23             continue;
24         }
25         node f=q.top();
26         while(nowt+f.t2<=x.t1){
27             q.pop();
28             nowt=nowt+f.t2;
29             printf("%d %d\n",f.num,nowt);
30             if(q.empty()) break;
31             f=q.top();
32         }
33         if(q.empty()){
34             nowt=x.t1;
35             q.push(x);
36             continue;
37         }
38         f.t2-=x.t1-nowt;
39         q.pop();
40         q.push(f);
41         nowt=x.t1;
42         q.push(x); 
43     }
44     while(!q.empty()){
45         node f=q.top();
46         q.pop();
47         nowt=nowt+f.t2;
48         printf("%d %d\n",f.num,nowt);
49     }
50     return 0;
51 }

// HNOI2003 Day1t2