AtCoder Beginner Contest 137 D - Summer Vacation——————貪心，優先隊列

abc137_D - 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.

• $1≤N≤10^5$
• $1≤M≤10^5$
• $1≤A_i≤10^5$
• $1≤B_i≤10^4$

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

$N$ $M$
$A_1$ $B_1$
$A_2$ $B_2$
$⋮$
$A_N$ $B_N$

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

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每天只能領取一個工作，不過每天可以做的工作可以有多個
比如樣例$1$，第一天領取$4\ 3$，在第四天可以得到3的獎勵，第二天領取$2\ 2$，第三天可以獲得2的獎勵

我們可以知道，在第$i$天，不能領取工作時長超過$m-i+1$的工作，於是我們就可以在這些滿足條件的工作中選擇獎勵最高的。
倒着想就行了，在倒數第$i$天，我能選擇的工作是：工作時長不超過$i$的，在這些滿足條件的工作中選擇獎勵最高的。
用個優先隊列維護一下，不過得先排序，按照天數遞增排序

貪心，思維，優先隊列

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5+7;
typedef long long ll;

struct node{
        ll a,b;
}z[MAXN];

bool cmp(node x,node y) {
        return x.a<y.a;
}

int main() {
        int n,m;
        while(cin>>n>>m) {
                for(int i=0;i<n;++i)
                        cin>>z[i].a>>z[i].b;
                sort(z,z+n,cmp);
                ll ans = 0;
                int id = 0;
                priority_queue<ll> que;
                for(int i=1;i<=m;++i) {
                        while(z[id].a<=i && id<n) {
                                que.push(z[id].b);
                                id++;
                        }
                        if(que.empty()) continue;
                        ans += que.top();
                        que.pop();
                }
                cout<<ans<<endl;
        }
        return 0;
}