I just wonder if it was a common misinterpretation or it's only me, the problem says
If Ti=0, the i-th item is a pull-tab can; if you obtain it, you get a happiness of Xi.
If Ti=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 Xi.
If Ti=2, the i-th item is a can opener; it can be used against at most Xi cans.
for Ti=1, my understanding is I can use it with a can opener, but I don't see anywhere in the statement where it says I can use it with 0 happiness.
Atcoder says I need to carefully read the problem statement and I'm the only one who didn't understand the problem.
I understand I can use any Ti=0, I understand I can use any Ti=2, but it's not clear if Ti=1 I still can use it with 0 happiness.
https://atcoder.jp/contests/abc312/tasks/abc312_f