Hdu4705 tree dp

The number of any three points in the tree is not in a single path. Number of route schemes composed of three points S = C (n, 3) = n * (n-1) * (n-2) 6. using S minus the number of three points in a path is the answer. The enumeration center point is the second point. in its K-th sub-tree, the question is: any three points on the tree are not in a path, and how many of these three points are counted. Number of route schemes composed of three points S = C (n, 3) = n * (n-1) * (n-2) /6. the answer is to subtract the number of three points in one path from S. The enumeration center is used as the second vertex, and a vertex is found on its K-th subtree as the first vertex. in the rest of the nodes that have not been enumerated, find a vertex as the third vertex. Multiply the number. In this way, the problem becomes the subnode of the statistics node, simple tree-like dp. # Include # Include # Include # Include # Include # Pragma comment (linker, "/STACk: 102400000,102400000") using namespace std; vector Tree [100005]; typedef long ll; ll ans, ll n; ll dfs (int k, int fa =-1) {ll sum = 0, son = 0; for (int I = 0; I > N) {for (int I = 1; I <= n; I ++) tree [I]. clear (); for (int I = 1; I