http://lightoj.com/volume_showproblem.php?problem=1094 Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in the t…

1094 - Farthest Nodes in a Tree   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undi…

Farthest Nodes in a Tree Time Limit: 2000MS Memory Limit: 32768KB 64bit IO Format: %lld & %llu Submit Status Description Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weigh…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1094 Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in…

1094 - Farthest Nodes in a Tree PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undire…

1094 - Farthest Nodes in a Tree problem=1094" style="color:rgb(79,107,114)"> problem=1094&language=english&type=pdf" style="color:rgb(79,107,114)">PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Li…

题目链接,密码:hpu Description Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in the tree whose distance is maximum a…

Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in the tree whose distance is maximum amongst all nodes. Input…

http://lightoj.com/volume_showproblem.php?problem=1094 树的直径是指树的最长简单路. 求法: 两遍BFS :先任选一个起点BFS找到最长路的终点,再从终点进行BFS,则第二次BFS找到的最长路即为树的直径: 原理: 设起点为u,第一次BFS找到的终点v一定是树的直径的一个端点 证明: 1) 如果u 是直径上的点,则v显然是直径的终点(因为如果v不是的话,则必定存在另一个点w使得u到w的距离更长,则于BFS找到了v矛盾)2) 如果u不是直径上的…

树上最远点对(树的直径) 做法1:树形dp 最长路一定是经过树上的某一个节点的. 因此: an1[i],an2[i]分别表示一个点向下的最长链和次长链,次长链不存在就设为0:这两者很容易求 an3[i]表示i为根的子树中的答案:an3[u]=max(max{an3[v]}(v是u的子节点),an1[u]+an2[u]) #include<cstdio> #include<cstring> #include<queue> #include<algorithm>…