Work Scheduling
- Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
- Total Submission(s): 34 Accepted Submission(s): 3
Description
Farmer John has so very many jobs to do! In order to run the farm efficiently, he must make money on the jobs he does, each one of which takes just one time unit.
His work day starts at time 0 and has 1,000,000,000 time units (!). He currently can choose from any of N (1 ≤ N ≤ 100,000) jobs conveniently numbered 1..N for work to do. It is possible but extremely unlikely that he has time for all Njobs since he can only work on one job during any time unit and the deadlines tend to fall so that he can not perform all the tasks.
Job i has deadline Di (1 ≤ Di ≤ 1,000,000,000). If he finishes job i by then, he makes a profit of Pi (1 ≤ Pi ≤ 1,000,000,000).
What is the maximum total profit that FJ can earn from a given list of jobs and deadlines? The answer might not fit into a 32-bit integer.
Input
Multipel test cases. For each case :
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains two space-separated integers: Di and Pi
Output
For each case, output one line : A single number on a line by itself that is the maximum possible profit FJ can earn.
Sample Input
3
2 10
1 5
1 7
Sample Output
17
Hint
Complete job 3 (1, 7) at time 1 and complete job 1 (2, 10) at time 2 to maximize the earnings (7 + 10 → 17).
Source
USACO 2009 Open
1-100000000個機關時間内可以做n項工作，每項工作需要1個機關時間，截止時間為Di，價值為Pi。
運用貪心，将時間從小到大排序，取價值最大的工作。
用優先隊列
1 #include<bits/stdc++.h>
2 #define man 100006
3 #define ll long long
4 using namespace std;
5 ll n,ans;
6 priority_queue<ll,vector<ll>,greater<ll> >in；///優先隊列
7 typedef struct W{
8 ll ti,cost;
9 };
10 W work[man];
11
12 bool cmp(W a, W b)
13 {
14 return a.ti<b.ti;
15 }
16
17 int main()
18 {
19 while( ~scanf("%lld",&n)){
20 ll x,y,si=0;
21 while(!in.empty())
22 in.pop();
23 for(int i=1;i<=n;i++){
24 scanf("%lld%lld",&x,&y);
25 work[i].ti=x;
26 work[i].cost=y;
27 }
28 sort( work+1, work+1+n, cmp);
29 ans=0;
30 for(int i=1;i<=n;i++){
31 if(si<work[i].ti){
32 si++;
33 in.push(work[i].cost);
34 ans+=work[i].cost;
35 }
36 else{
37 if(work[i].cost<=in.top())
38 continue;
39 ans-=in.top();
40 ans+=work[i].cost;
41 in.pop();
42 in.push(work[i].cost);
43 }
44 }
45
46 printf("%lld\n",ans);
47 }
48 return 0;
49 } View Code