Labyrinth Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 4004   Accepted: 1504 Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them eithe…

Labyrinth 题目链接: http://acm.hust.edu.cn/vjudge/contest/130510#problem/E Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or f…

题目连接 http://poj.org/problem?id=1383 Labyrinth Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or free. There is also a litt…

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…

Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or free. There is also a little hook on the floor in the center of every fr…

Cow Marathon Description 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 of farms and…

The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or free. There is also a little hook on the floor in the center of every free block. Th…

大致题意:在某个点派出两个点去遍历全部的边,花费为边的权值,求最少的花费 思路:这题关键好在这个模型和最长路模型之间的转换.能够转换得到,全部边遍历了两遍的总花费减去最长路的花费就是本题的答案,要思考.并且答案和派出时的起点无关 求最长路两遍dfs或bfs就可以,从随意点bfs一遍找到最长路的一个终点,再从这个终点bfs找到起点 //1032K 79MS C++ 1455B #include<cstdio> #include<iostream> #include<cstrin…

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…

题面 Labyrinth Time Limit: 2000MS Memory Limit: 32768K Total Submissions: 4997 Accepted: 1861 Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either…