1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013-11-17 14:30:29 4 File Name :E:\2013ACM\专题学习\树的分治\POJ1741.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <str…

题目描述 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…

D Tree Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others) Total Submission(s): 1687    Accepted Submission(s): 263 Problem Description There is a skyscraping tree standing on the playground of Nanjing University of…

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 e…

http://poj.org/problem?id=1741   题意:一棵n个点的树,每条边有距离v,求该树中距离小于等于k的点的对数.   dis[y]表示点y到根x的距离,v代表根到子树根的距离: 那么不在同一棵子树中的两点i.j之间的距离为dis[i]+dis[j]: ①    设得到这个距离的时间复杂度为O(w): 如果我们层层如此递归即可得到所有的点对数量,可以证明复杂度为O(logn*w);   因为n的范围为(n<=10000)所以我们需要w与n近似:   那么此时问题转化为了如…

Tree Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 23380   Accepted: 7748 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…

昨天学了下树分治,今天补这道题,还是太不熟练了,写完之后一直超时.后来查出好多错= =比如v,u写倒了,比如+写成了取最值,比如....爆int...查了两个多小时的错..哭...(没想到进首页了 http://hzwer.com/6107.html 大神博客,代码清晰,照着这个改的 逆元预处理之前是没有见过的,学习了. #include <iostream> #include <cstdio> #include <cstring> #include <algor…

Tree   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…

给定一棵树,边上有权值,要统计有多少对点路径的权值和<=k 分治算法在树的路径中的应用 这个论文里面有分析. 任意两点的路径,要么过根结点,要么在子树中.如果在子树中,那么只要递归处理就行了. 所以只要统计过根结点的对数. 所以只要算出任意结点到根结点的距离,用dist数组存下来,然后排序dist数组,就可以在O(n)的时间内算出有多少对点满足条件,具体见counts函数 但是counter函数计算时,会把路径都在子树中的情况也给加进来,所以需要再对相应的子树进行counts函数,然后减去. 然…

Tree Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21357   Accepted: 7006 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 in…