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…

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: 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…

<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…

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…

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 twice. Give…

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 twice. Give…

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…

树的直径: 利用了树的直径的一个性质:距某个点最远的叶子节点一定是树的某一条直径的端点. 先从任意一顶点a出发,bfs找到离它最远的一个叶子顶点b,然后再从b出发bfs找到离b最远的顶点c,那么b和c之间的距离就是树的直径. 用dfs也可以. 模板: ; int head[N]; int dis[N]; bool vis[N]; ,b,mxn=; struct edge { int to,w,next; }edge[N]; void add_edge(int u,int v,int w) { e…

Roads in the North POJ - 2631 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…

题目链接:http://poj.org/problem?id=2631 求树的直径模板. 定理: 树上任意一个点的在树上的最长路一定以树的直径的两端点其中一点结束. 做法: 两边bfs,第一次先找到node(树的直径的两端点其中一个),再一次求node的最长路所结束的点t node->t就是树的直径 #include <queue> #include <cstdio> #include <cstring> #include <iostream> #in…

http://poj.org/problem?id=2631 树的直径裸题 dfs/bfs均可 /* dfs */ #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #include <string> using namespace std; ; #define yxy getchar() , p…

树的直径:从随意一点出发,BFS找到最远的距离,然后在从该点出发BFS找到最远的距离 #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <queue> #include <cmath> #include <deque> #include <vector> using namespace s…

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=…

题目链接:http://poj.org/problem?id=1985 After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has committed to create a bovine marathon for his cows to run. The marathon route will include a pair o…

POJ 1724 ROADS Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12766   Accepted: 4722 Description N cities named with numbers 1 ... N are connected with one-way roads. Each road has two parameters associated with it : the road length and…

问题: 4个标记为1,2,3,4的节点构成自由树(算法导论里的定义,连接着,无环,无向的图),一共有多少种构造方法?如果N个节点呢? 解决方法: 4个节点可以通过穷举的方式得到答案,一共有16中方式. 第一类构造方式,取一个节点做中心,剩余三个节点与其相连,一共4种(每个节点做一次中心). 第二类构造方式,四个节点连成一条线,可以看成个排列,第一个有4种取法,第二个有3种,依次类推.但是因为例如1234与4321结构上是一样的, 所以总的排列数除以2才是总共的构造数,即 $\frac{4!}{2…

Cow Marathon Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 4185   Accepted: 2118 Case Time Limit: 1000MS Description After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has com…

树的直径:树上的最长简单路径. 求解的方法是bfs或者dfs.先找任意一点,bfs或者dfs找出离他最远的那个点,那么这个点一定是该树直径的一个端点,记录下该端点,继续bfs或者dfs出来离他最远的一个点,那么这两个点就是他的直径的短点,距离就是路径长度.具体证明见http://www.cnblogs.com/wuyiqi/archive/2012/04/08/2437424.html 其实这个自己画画图也能理解. POJ 1985 题意:直接让求最长路径. 可以用dfs也可以用bfs bfs代…

POJ 2152 fire / SCU 2977 fire(树型动态规划) Description Country Z has N cities, which are numbered from 1 to N. Cities are connected by highways, and there is exact one path between two different cities. Recently country Z often caught fire, so the governm…

求树直径的方法在此转载一下大佬们的分析: 可以随便选择一个点开始进行bfs或者dfs,从而找到离该点最远的那个点(可以证明,离树上任意一点最远的点一定是树的某条直径的两端点之一:树的直径:树上的最长简单路径).再从找到的点出发,找到据该点的最远点,那么这两点就确定了树的一条直径,两点间距即为所求距离. 无意中看到一道水题,也就是POJ 1383题目中给出了一个无环的迷宫,求出其中最长的一条路我们知道无环图本质上可以认为就是树,所以此题完全可以使用树的最长链算法 即:随便从某个节点C开始DFS或B…

POJ.3321 Apple Tree ( DFS序 线段树 单点更新 区间求和) 题意分析 卡卡屋前有一株苹果树,每年秋天,树上长了许多苹果.卡卡很喜欢苹果.树上有N个节点,卡卡给他们编号1到N,根的编号永远是1.每个节点上最多结一个苹果.卡卡想要了解某一个子树上一共结了多少苹果. 现在的问题是不断会有新的苹果长出来,卡卡也随时可能摘掉一个苹果吃掉.你能帮助卡卡吗? 前缀技能 边表存储树 DFS时间戳 线段树 首先利用边表将树存储下来,然后DFS打上时间戳.打上时间戳之后,我们就知道书上节点对…

http://poj.org/problem?id=1985 题意:给出树,求最远距离. 题意: 树的直径. 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…

Cow Marathon Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 7536   Accepted: 3559 Case Time Limit: 1000MS Description After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has com…

Roads in the North Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2941   Accepted: 1447 Description Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such…

Cow Marathon Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 5496   Accepted: 2685 Case Time Limit: 1000MS Description After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has com…