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 ${\mathrm{Bi}}^{}$ after ${\mathrm{Ai}}^{}$ 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.
• $\mathrm{1\le N\le 105}$
• $\mathrm{1\le M\le 105}$
• $\mathrm{1\le Ai\le 105}$
• $\mathrm{1\le Bi\le 104}$

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Input

Input is given from Standard Input in the following format:

$N$  $M$
${\mathrm{A1}}^{}$ ${\mathrm{B1}}^{}$
${\mathrm{A2}}^{}$ ${\mathrm{B2}}^{}$
$⋮⋮$
${\mathrm{AN}}^{}$ ${\mathrm{BN}}^{}$

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

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
struct wzy
{
    int day,money;
}p[maxn];
bool cmp(wzy u,wzy v)
{
    return u.day<v.day;
}
int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++)
        cin>>p[i].day>>p[i].money;
    sort(p,p+n,cmp);
    priority_queue<int>que;
    int ans=0;
    int pos=0;
    for(int i=1;i<=m;i++)
    {
        while(p[pos].day<=i&&pos<n)
            que.push(p[pos++].money);
        if(!que.empty())
            ans+=que.top(),que.pop();
    }
    cout<<ans<<endl;
    return 0;
}