POJ [P2631] Roads in the North】的更多相关文章

树的直径 树的直径求法: 任取一点u,找到树上距u最远的点s 找到树上距s点最远的点t,s->t的距离即为所求 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <cstdlib> #include <cmath> #include <queue> using namespace std; in…
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…
题目连接 http://poj.org/problem?id=2631 Roads in the North 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…
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…
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: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…
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 not pass through some other village…
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 not pass through some other village…
<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…