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.
  1≤N≤10⁵
  1≤M≤10⁵
  1≤A_i≤10⁵
  1≤B_i≤10⁴
Input
  Input is given from Standard Input in the following format:
  N M
  A₁ B₁
  A₂ B₂
   ⋮
  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.

题意

  给定n份工作，第 i 工作需要工作 A_i 天才能收入 B_i ，你每天可以开始一份工作，但是每份工作只能做一次，问你 m 天的最大收入

思路

  首先只要这个工作的需要时间 <= m，很明显，最后一天一定是在耗时为1的里面找最优解，而最后两天是在耗时为 1 和 2 里找最优解，以此类推……
  开一个优先队列，从 1~m 暴一遍即可

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f3f;
const int maxn = 100100;
vector<int> p[maxn];	//分类存储
priority_queue<int> q;	//找最优情况
int main(void)
{
    int n,m;
    int a,b;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&a,&b);
        p[a].push_back(b);
    }
    int ans = 0;
    for(int i=1;i<maxn;i++) sort(p[i].begin(),p[i].end());	//排序
    for(int i=1;i<=m;i++){
        for(int j=0;j<p[i].size();j++){
            q.push(p[i][j]);	//入队
        }
        if(!q.empty()){
            ans += q.top();	//找最优
            q.pop();	//出队
        }
    }
    printf("%d\n",ans);
    return 0;
}