P5851 [USACO19DEC] Greedy Pie Eaters P

Farmer John has $M$ cows, conveniently labeled $1 \ldots M$, who enjoy the occasional change of pace from eating grass. As a treat for the cows, Farmer John has baked $N$ pies ($1 \leq N \leq 300$), labeled $1 \ldots N$. Cow $i$ enjoys pies with labels in the range $[l_i, r_i]$ (from $l_i$ to $r_i$ inclusive), and no two cows enjoy the exact same range of pies. Cow $i$ also has a weight, $w_i$, which is an integer in the range $1 \ldots 10^6$. Farmer John may choose a sequence of cows $c_1,c_2,\ldots, c_K,$ after which the selected cows will take turns eating in that order. Unfortunately, the cows don't know how to share! When it is cow $c_i$'s turn to eat, she will consume all of the pies that she enjoys --- that is, all remaining pies in the interval $[l_{c_i},r_{c_i}]$. Farmer John would like to avoid the awkward situation occurring when it is a cows turn to eat but all of the pies she enjoys have already been consumed. Therefore, he wants you to compute the largest possible total weight ($w_{c_1}+w_{c_2}+\ldots+w_{c_K}$) of a sequence $c_1,c_2,\ldots, c_K$ for which each cow in the sequence eats at least one pie.

The first line contains two integers $N$ and $M(1\leq M\leq \frac{N(N+1)}{2})$ The next $M$ lines each describe a cow in terms of the integers $w_i$,$l_i$, and $r_i$.

Print the maximum possible total weight of a valid sequence.

In this example, if cow $1$ eats first, then there will be nothing left for cow $2$ to eat. However, if cow $2$ eats first, then cow $1$ will be satisfied by eating the second pie only. Test cases $2-5$ satisfy $N\leq50$ and $M\leq 20$. Test cases $6-9$ satisfy $N\leq 50$.