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).

My submission:http://codeforces.net/contest/808/submission/27174539