用vc++实现用最短路径查找哈弗曼树。n个权值构成n棵二叉树的集合F={T1, T2, …, Tn}，其中每棵二叉树Ti中只有一个带树为Ti的根结点。-Achieved with the vc++ to find the shortest path tree Havermann. n-weight constitutes a collection of n binary trees F = (T1, T2, ..., Tn), which is only a Ti binary Each species of tree with the root node for the Ti.