I'm sorry to ask a foolish problem about my submission.So I share a way to solve 808E by ternary searching.
As we can see,W[i] = 1, 2, 3. We sort the Souvenirs which have the same W[i] by their C[i]. Then we know we will just choose a prefix for each W[i].
If we get X Souvenirs that W[i] = 3,we remain (m - 3X) to get Souvenirs that W[i] = 1, 2. Let's define F[i][x] the prefix sum of Souvenirs that W[i] = x. If we get Y Souvenirs that W[i] = 2,our cost is F[X] + F[Y] + F[m - 3X - 2Y] (m-3X-2Y>=0).
We can try every X and use ternary searching to find the best Y.The complexity is O(nlogn).