Summer Vacation

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.

Input
Input is given from Standard Input in the following format:

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
题意：
给出n（可选工作方案），m（天数）
接下来的n 行 An Bn 表示在今天做此工作 An 天后可以领取工资Bn；
分析：
此题用到贪心思想，开始分析时认为是背包问题，看完题解后明白可以使用贪心的思想来解此题。m天数，可以倒着来算，将满足在倒着数1,2 ,3~~m天工作时最大工资求和即可。可利用优先队列将满足工作天数的工资入队，再取出队首最大工资求和并去除即可；
代码

#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
struct node{
	int a,b;
}s[maxn];
bool cmp(node a,node b)
{
	return a.a<b.a;
}
int main()
{
	int n,m;
	int ant=0;
	ll ans=0;
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
	{
		scanf("%d%d",&s[i].a,&s[i].b);
	}
	priority_queue<int>q;
	sort(s,s+n,cmp);//以发工资天数排序 
	for(int i=1;i<=m;i++)
	{
		while(s[ant].a<=i&&ant<n)//将满足条件工作工资入队 
		{
			q.push(s[ant].b);
			ant++;
		}
		if(!q.empty())//取出队首最大工资 
		{
			ans+=q.top();
			q.pop();
		}
	}
	printf("%lld\n",ans); 
	return 0;
 }