Score : $500$ points

问题描述

有 $N$ 件物品。
其中每一件是易拉罐、普通罐头或开罐器之一。
第 $i$ 个物品由一个整数对 $(T_i, X_i)$ 描述如下：

• 如果 $T_i = 0$，则第 $i$ 个物品是一个易拉罐；如果你获取它，你会得到幸福度为 $X_i$。
• 如果 $T_i = 1$，则第 $i$ 个物品是一个普通罐头；如果你获取它并用开罐器打开它，你会得到幸福度为 $X_i$。
• 如果 $T_i = 2$，则第 $i$ 个物品是一个开罐器；它可以最多用来打开 $X_i$ 个罐头。

在 $N$ 件物品中选取 $M$ 件，求能获得的最大总幸福度。

以上为通义千问 qwen-max 翻译，仅供参考。

Problem Statement

There are $N$ items.
Each of these is one of a pull-tab can, a regular can, or a can opener.
The $i$-th item is described by an integer pair $(T_i, X_i)$ as follows:

• If $T_i = 0$, the $i$-th item is a pull-tab can; if you obtain it, you get a happiness of $X_i$.
• If $T_i = 1$, the $i$-th item is a regular can; if you obtain it and use a can opener against it, you get a happiness of $X_i$.
• If $T_i = 2$, the $i$-th item is a can opener; it can be used against at most $X_i$ cans.

Find the maximum total happiness that you get by obtaining $M$ items out of $N$.

Constraints

• $1 \leq M \leq N \leq 2 \times 10^5$
• $T_i$ is $0$, $1$, or $2$.
• $1 \leq X_i \leq 10^9$
• All input values are integers.

Input

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

$N$ $M$

$T_1$ $X_1$

$T_2$ $X_2$

$\vdots$

$T_N$ $X_N$

Output

Print the answer as an integer.

Sample Input 1

8 4
0 6
0 6
1 3
1 5
1 15
2 1
2 10
2 100

Sample Output 1

27

If you obtain the $1$-st, $2$-nd, $5$-th, and $7$-th items, and use the $7$-th item (a can opener) against the $5$-th item, you will get a happiness of $6 + 6 + 15 = 27$.
There are no ways to obtain items to get a happiness of $28$ or greater, but you can still get a happiness of $27$ by obtaining the $6$-th or $8$-th items instead of the $7$-th in the combination above.

Sample Input 2

5 5
1 5
1 5
1 5
1 5
1 5

Sample Output 2

0

Sample Input 3

12 6
2 2
0 1
0 9
1 3
1 5
1 3
0 4
2 1
1 8
2 1
0 1
0 4

Sample Output 3

30

update @ 2024/3/10 08:54:03