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][3] + F[Y][2] + F[m-3X-2Y][1] (m-3X-2Y>=0).

We can try every X and use ternary searching to find the best Y.The complexity is O(nlogn).