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. Wri…

写的第一道点分治的题目,权当认识点分治了. 点分治,就是对每条过某个点的路径进行考虑,若路径不经过此点,则可以对其子树进行考虑. 具体可以看menci的blog:点分治 来看一道例题:POJ 1741 Tree 题目大意:扔给你一颗有权无根树,求有多少条路径的长度小于k: 解题思路:先找出重心,用一次dfs处理出每个点到根的距离dis,然后将dis[]排序,用O(n)的复杂度处理出"过根且长度小于等于k的路径数目",删除根节点,对于每棵子树重复上述操作. 注意要去重: 像上面这样一个图…

n k n个节点的一棵树 k是距离 求树上有几对点距离<=k; #include<stdio.h> #include<string.h> #include<algorithm> #include<vector> using namespace std; #define MAXN 100100 int head[MAXN]; bool vis[MAXN]; struct edg { int to,next,w; }x[MAXN]; int cnt,ans…

题目链接:http://poj.org/problem?id=1741 题意: 给定一棵包含$n$个点的带边权树,求距离小于等于K的点对数量 题解: 显然,枚举所有点的子树可以获得答案,但是朴素发$O(n^2logn)$算法会超时, 利用树的重心进行点分治可以将$O(n^2logn)$的上界优化为近似$O(nlogn)$ 足以在1000ms的测试时间内通过 具体原理参考注释 #include<iostream> #include<map> #include<string>…

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…

                                                                                             POJ 1741 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…

POJ 1741. Tree Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 34141   Accepted: 11420 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…

Tree Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 18205   Accepted: 5951 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…

poj 1741 Tree(树的点分治) 给出一个n个结点的树和一个整数k,问有多少个距离不超过k的点对. 首先对于一个树中的点对,要么经过根结点,要么不经过.所以我们可以把经过根节点的符合点对统计出来.接着对于每一个子树再次运算.如果不用点分治的技巧,时间复杂度可能退化成\(O(n^2)\)(链).如果对于子树重新选根,找到树的重心,就一定可以保证时间复杂度在\(O(nlogn)\)内. 具体技巧是:首先选出树的重心,将重心视为根.接着计算出每个结点的深度,以此统计答案.由于子树中可能出现重复…

http://poj.org/problem? id=1741 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 va…