CF960E Alternating Tree

Given a tree with $ n $ nodes numbered from $ 1 $ to $ n $ . Each node $ i $ has an associated value $ V_i $ . If the simple path from $ u_1 $ to $ u_m $ consists of $ m $ nodes namely $ u_1 \rightarrow u_2 \rightarrow u_3 \rightarrow \dots u_{m-1} \rightarrow u_{m} $ , then its alternating function $ A(u_{1},u_{m}) $ is defined as $ A(u_{1},u_{m}) = \sum\limits_{i=1}^{m} (-1)^{i+1} \cdot V_{u_{i}} $ . A path can also have $ 0 $ edges, i.e. $ u_{1}=u_{m} $ . Compute the sum of alternating functions of all unique simple paths. Note that the paths are directed: two paths are considered different if the starting vertices differ or the ending vertices differ. The answer may be large so compute it modulo $ 10^{9}+7 $ .

N/A

N/A

Consider the first example. A simple path from node $ 1 $ to node $ 2 $ : $ 1 \rightarrow 2 $ has alternating function equal to $ A(1,2) = 1 \cdot (-4)+(-1) \cdot 1 = -5 $ . A simple path from node $ 1 $ to node $ 3 $ : $ 1 \rightarrow 3 $ has alternating function equal to $ A(1,3) = 1 \cdot (-4)+(-1) \cdot 5 = -9 $ . A simple path from node $ 2 $ to node $ 4 $ : $ 2 \rightarrow 1 \rightarrow 4 $ has alternating function $ A(2,4) = 1 \cdot (1)+(-1) \cdot (-4)+1 \cdot (-2) = 3 $ . A simple path from node $ 1 $ to node $ 1 $ has a single node $ 1 $ , so $ A(1,1) = 1 \cdot (-4) = -4 $ . Similarly, $ A(2, 1) = 5 $ , $ A(3, 1) = 9 $ , $ A(4, 2) = 3 $ , $ A(1, 4) = -2 $ , $ A(4, 1) = 2 $ , $ A(2, 2) = 1 $ , $ A(3, 3) = 5 $ , $ A(4, 4) = -2 $ , $ A(3, 4) = 7 $ , $ A(4, 3) = 7 $ , $ A(2, 3) = 10 $ , $ A(3, 2) = 10 $ . So the answer is $ (-5) + (-9) + 3 + (-4) + 5 + 9 + 3 + (-2) + 2 + 1 + 5 + (-2) + 7 + 7 + 10 + 10 = 40 $ . Similarly $ A(1,4)=-2, A(2,2)=1, A(2,1)=5, A(2,3)=10, A(3,3)=5, A(3,1)=9, A(3,2)=10, A(3,4)=7, A(4,4)=-2, A(4,1)=2, A(4,2)=3 , A(4,3)=7 $ which sums upto 40.