The important observation here is that we are only concerned with the parity of no. of bits on any path, as mentioned in the editorial. Thus, we need to only look at the no. of set bits in a number, and not which bits are set.

Then we can impose some restrictions on the parity of bits on each path in the input. Any edge with non negative value can also be viewed as a path of size 1. Hence, we can try to impose all these restrictions and see if these give us a valid combination. Then, we will work with the negative weighted edges and obtain their values in the same way.

To store the parity of bits in any path, we are using DSU (but it is probably not required if we root our tree beforehand). To obtain the parity between any two nodes, we can xor the parity of each path from the root of the tree. Since, if any edge occurs in both the paths, it will not be on the simple path between the two nodes and hence get cancelled. If at any point, the conditions are not satisfied or are contradictory, we know that the required tree is not possible. Finally to obtain the weights of the negative edges, we can check for the required parity between the two nodes, and set our answer as 0 or 1. Here is my solution, feel free to ask any queries.

https://mirror.codeforces.com/contest/1615/submission/269522430

The solution can also be viewed as a variant of bipartite matching. Each node is colored either 0 or 1, depending on the parity of path between the node and root.

are you not tryna reach expert lmao

then ask robezh for more problems smh

surely he can offer you some 603 probs