Description
Give a tree with n vertices,each edge has a length(positive integer less than 1001).
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Input
The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l.
The last test case is followed by two zeros.
The last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
Sample Input
5 4 1 2 3 1 3 1 1 4 2 3 5 1 0 0
Sample Output
8
题目意思为给一个带边权的树，求满足dist（u，v）<=k 的（u，v）的对数
解决方法，分治法，树上的点分治，可以考虑u和v分别过跟节点，对于root，求出所有子节点到它的距离 d[i]，然后统计d[i]+d[j]<=k 的对数，这里可以用单调队处理,然后我们就递归处理root的孩子，正因为这样，我们多算了处于同一个子树的，故要减去。
其实这些都不是重要的，这些算法网上都说的很清楚，反而是细节没有说，就是每次处理的时候，都要求一次root，root就是树的重心，用dp求，不然还是会超时的