Atcoder ABC137D：Summer Vacation（贪心）

D - Summer Vacation

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

• All values in input are integers.
• 1≤N≤10⁵
  
• 1≤M≤10⁵
  
• 1≤A_i≤10⁵
  
• 1≤B_i≤10⁴
  

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Input

Input is given from Standard Input in the following format:

N  M
A1 B1
A2 B2
⋮⋮
AN BN

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

• Take and complete the first job today. You will earn the reward of 3 after four days from today.
• Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

题意

一共有N个任务和M天，一个人一天只能做一个任务，做完任务之后可以在第A_i天拿到B_i的工资，问M天内最多可以拿到多少工资

思路

贪心，按照领取工资间隔升序排序

设置一个优先队列用来储存当前可以做的任务并且可以在M天内拿到工资的任务的钱数，每次取可以拿到的最大值

代码

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 struct wzy
12 {
13     int day,money;
14 }p[maxn];
15 bool cmp(wzy u,wzy v)
16 {
17     return u.day<v.day;
18 }
19 int main(int argc, char const *argv[])
20 {
21     ios::sync_with_stdio(false);
22     cin.tie(0);
23     int n,m;
24     cin>>n>>m;
25     for(int i=0;i<n;i++)
26         cin>>p[i].day>>p[i].money;
27     sort(p,p+n,cmp);
28     priority_queue<int>que;
29     int ans=0;
30     int pos=0;
31     for(int i=1;i<=m;i++)
32     {
33         while(p[pos].day<=i&&pos<n)
34             que.push(p[pos++].money);
35         if(!que.empty())
36             ans+=que.top(),que.pop();
37     }
38     cout<<ans<<endl;
39     return 0;
40 }