POJ 2631 Roads in the North(树的直径) http://poj.org/problem? id=2631 题意: 有一个树结构, 给你树的全部边(u,v,cost), 表示u和v两点间有一条距离为cost的边. 然后问你该树上最远的两个点的距离是多少?(即树的直径) 分析: 对于树的直径问题, <<算法导论>>(22 2-7)例题有说明. 详细解法: 首先从树上随意一个点a出发, (BFS)找出到这个点距离最远的点b. 然后在从b点出发(BFS)找到距离b…

题目大意:给定一棵 N 个节点的边权无根树,求树的直径. 代码如下 #include <cstdio> #include <algorithm> using namespace std; const int maxn=1e4+10; struct node{ int nxt,to,w; }e[maxn<<1]; int tot=1,head[maxn]; inline void add_edge(int from,int to,int w){ e[++tot].nxt=…

Roads in the North Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2389   Accepted: 1173 Description Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such…

Codeforces 题目传送门 & 洛谷题目传送门 其实是一道还算一般的题罢--大概是最近刷长链剖分,被某道长链剖分与直径结合的题爆踩之后就点开了这题. 本题的难点就在于看出一个性质:最长路径的其中一个端点一定是直径的某一个端点. 证明:首先我们找出原树的一个直径,如果直径上标记边的个数为偶数那显然这条直径就是最优解,符合题意,否则我们假设我们找出的直径为 \(AB\),我们已经找出了一条符合要求的路径 \(CD\),下证我们总可以通过调整 \(CD\) 的端点,找出一条以 \(A\) 或 \…

Roads in the North Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2359   Accepted: 1157 Description Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such…

题目链接:http://poj.org/problem?id=2631 Description Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such that there is only one route from a village to a village that does…

Roads in the North Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4513   Accepted: 2157 Description Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such…

<题目链接> 题目大意:求一颗带权树上任意两点的最远路径长度. 解题分析: 裸的树的直径,可由树形DP和DFS.BFS求解,下面介绍的是BFS解法. 在树上跑两遍BFS即可,第一遍BFS以任意点为起点,此时得到的离它距离最远的点为树的直径上的端点之一,然后再以这个端点为起点,跑一遍BFS,此时离它最远的点为树直径的另一个端点,同时,它们之间的距离即为树的直径. #include<iostream> #include<cstdio> #include<algorit…

题意: 给定一棵树, 求树的直径. 分析: 两种方法: 1.两次bfs, 第一次求出最远的点, 第二次求该点的最远距离就是直径. 2.同hdu2196的第一次dfs, 求出每个节点到子树的最长距离和次长距离, 然后某个点的最长+次长就是直径. #include<stdio.h> #include<vector> #include<algorithm> #include<string.h> #include<iostream> using name…

poj1985 Cow Marathon 树的直径裸题 树的直径的一般求法: 任意一点为起点,dfs/bfs找出与它最远的点$u$ 以$u$为起点,dfs/bfs找出与它最远的点$v$ 则$d(u,v)$是一条直径 下面给出poj1985的code(poj2631自行修改) #include<iostream> #include<cstdio> #include<cstring> #define re register using namespace std; int…