Roads in the North（POJ 2631 DFS）

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Roads in the North（POJ 2631 DFS）

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…

题解报告：poj 2631 Roads in the North（最长链）

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…

POJ 2631 Roads in the North （树的直径）

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…

Roads in the North （树的直径）

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…

Hopscotch（POJ 3050 DFS）

Hopscotch Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2845 Accepted: 1995 Description The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows creat…

poj 2631 Roads in the North （自由树的直径）

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…

深度搜索DFS-Lake Counting（POJ NO.2386）

题目链接POJ NO.2386 解题思路: 这个也是一个dfs 的应用,在书上的例子,因为书上的代码并不全,基本都是函数分块来写,通过这个题目也规范了代码,以后能用函数的就都用函数来实现吧.采用深度优先搜索,从任意的w开始,不断把邻接的部分用'.'代替,1次DFS后与初始这个w连接的所有w就全都被替换成'.',因此直到图中不再存在W为止,总共进行DFS的次数就是答案.8个方向对应8个状态转移,每个格子作为DFS的参数最多调用一次,因此时间复杂度为O(8nm)=O(nm). AC 代码: #inc…