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POJ 1741 Tree

时间:2015-01-12 23:50:04      阅读:222      评论:0      收藏:0      [点我收藏+]

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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. 

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. 

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

POJ 1741 Tree

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原文地址:http://www.cnblogs.com/chensunrise/p/4220030.html

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