Problem:
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 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

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.

Sample Input 2

5 3
1 2
1 3
1 4
2 1
2 3

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)


#include <iostream>
#include <queue>
#include <cstring>

using namespace std;
typedef long long LL;

int main()
{
    int N, M;
    cin >> N >> M;

    priority_queue<int>* jobs[100001];
    memset(jobs, 0, sizeof(priority_queue<int>*) * 100001);

    for (int i = 0; i < N; ++i) {
        int A, B;
        cin >> A >> B;
        if (A <= M) {
            if (!jobs[A]) {
                jobs[A] = new priority_queue<int>();
            }
            jobs[A]->push(B);
        }
    }

    LL ans;
    int r = 1; //! remaining days
    priority_queue<int> payableJobs;
    for (int r = 1; r <= M; ++r) {
        if (jobs[r]) {
            while (!jobs[r]->empty()) {
                payableJobs.push(jobs[r]->top());
                jobs[r]->pop();
            }
        }       
        if (!payableJobs.empty()) {
            ans += payableJobs.top();
            payableJobs.pop();
        }
    }

    cout << ans << endl;

    return 0;
}