感觉很生疏. https://leetcode.com/problems/word-ladder/…

一.基本思想 1)从图中的某个顶点V出发访问并记录: 2)依次访问V的所有邻接顶点: 3)分别从这些邻接点出发,依次访问它们的未被访问过的邻接点,直到图中所有已被访问过的顶点的邻接点都被访问到. 4)重复第3步,直到图中所有顶点都被访问完为止.   二.图的存储结构…

6-2 邻接表存储图的广度优先遍历(20 分) 试实现邻接表存储图的广度优先遍历. 函数接口定义: void BFS ( LGraph Graph, Vertex S, void (*Visit)(Vertex) ); 其中LGraph是邻接表存储的图,定义如下: /* 邻接点的定义 */ typedef struct AdjVNode *PtrToAdjVNode; struct AdjVNode{ Vertex AdjV; /* 邻接点下标 */ PtrToAdjVNode Next; /*…

试实现邻接表存储图的广度优先遍历. 函数接口定义: void BFS ( LGraph Graph, Vertex S, void (*Visit)(Vertex) ) 其中LGraph是邻接表存储的图,定义如下: /* 邻接点的定义 */ typedef struct AdjVNode *PtrToAdjVNode; struct AdjVNode{ Vertex AdjV; /* 邻接点下标 */ PtrToAdjVNode Next; /* 指向下一个邻接点的指针 */ }; /* 顶点表…

在上一篇文章我们用java演示了图的数据结构以及图涉及到的深度优先遍历算法,本篇文章将继续演示图的广度优先遍历算法.广度优先遍历算法主要是采用了分层的思想进行数据搜索.其中也需要使用另外一种数据结构队列,本篇文章为了使代码更加优雅,所有使用java中Linkedlist集合来进行模拟队列.因为该集合有在队列尾部添加元素和从队头取出元素的API. 算法思想: 1.先访问一个元素,然后放到队列中,并且标记已经访问过该元素. 2.然后判断队列是否为空,不为空则取出队头元素. 3.然后取出队头元素的第一…

广度优先遍历 广度优先遍历是非经常见和普遍的一种图的遍历方法了,除了BFS还有DFS也就是深度优先遍历方法.我在我下一篇博客里面会写. 遍历过程 相信每一个看这篇博客的人,都能看懂邻接链表存储图. 不懂的人.请先学下图的存储方法.在我的之前博客里. 传送门:图表示方法 然后我们如果有一个图例如以下: 节点1->3->NULL 节点2->NULL 节点3->2->4->NULL 节点4->1->2->NULL 这样我们已经知道这是一个什么图了. 如果我们…

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, Given:start = "hit…

广度优先遍历: 1.将起点s 放入队列Q(访问) 2.只要Q不为空,就循环执行下列处理 (1)从Q取出顶点u 进行访问(访问结束) (2)将与u 相邻的未访问顶点v 放入Q, 同时将d[v]更新为d[u] + 1 主要变量 M[n][n] 邻接矩阵,如果存在顶点i到顶点j 的边,则M[i][j] 为true Queue Q 记录下一个待访问顶点的队列 d[n] 将起点s 到个顶点i的最短距离记录在d[i]中. s无法到达i 时d[i] 为INFTY(极大值) #include<iostream>…

环境需求:JDK:1.8 jar:jgrapht-core-1.01.jar package edu; import org.jgrapht.experimental.dag.DirectedAcyclicGraph; import org.jgrapht.experimental.dag.DirectedAcyclicGraph.CycleFoundException; import org.jgrapht.graph.DefaultEdge; import org.jgrapht.trave…

Word Ladder II Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a timeEach intermediate word must exist in the dictionaryFor example, Given…

主要参考资料:数据结构(C语言版)严蔚敏   ,http://blog.chinaunix.net/uid-25324849-id-2182922.html   代码测试通过. package 图的建立与实现; import java.util.*; public class MGraph { final int MAXVEX = 100; final int INFINITY = 65535; int[] vexs = new int[MAXVEX]; //顶点表 int[][] arc =…

题目大意:一张图,问从起点到终点的最短时间是多少.方向转动也消耗时间. 题目分析:图的广度优先遍历... 代码如下: # include<iostream> # include<cstdio> # include<queue> # include<cstring> # include<algorithm> using namespace std; struct Node { int x,y,f,l,t; Node(int _x,int _y,in…

首先需要知道的是,图的深度优先遍历是一种类似于树的前序遍历方式,即选择一个入口节点,沿着这个节点一直遍历下去,直至所有节点都被访问完毕:如果说,图的深度优先遍历类似于树的前序遍历的话,那么图的广度优先遍历就类似于树的层序遍历了,即一层一层的去遍历节点,直至所有节点都被访问完毕. 不多说,直接上代码,会对部分代码添加说明. import java.util.*; public class Traverse{ public static void main(String[] args){ //接受输…

XDOJ324.图的广度优先遍历 问题与解答 问题描述 已知无向图的邻接矩阵,以该矩阵为基础,给出广度优先搜索遍历序列,并且给出该无向图的连通分量的个数.在遍历时,当有多个点可选时,优先选择编号小的顶点.(即从顶点1开始进行遍历) 输入格式 第一行是1个正整数,为顶点个数n(n<100),顶点编号依次为1,2,-,n.后面是邻接矩阵,n行n列. 输出格式 共2行.第一行输出为无向图的广度优先搜索遍历序列,输出为顶点编号,顶点编号之间用空格隔开:第二行为无向图的连通分量的个数. 样例输入 6 0…

Word Ladder Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example,…

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, Given:start = "hit…

题目: Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, Given: start =…

原题地址 既然是求最短路径,可以考虑动归或广搜.这道题对字典直接进行动归是不现实的,因为字典里的单词非常多.只能选择广搜了. 思路也非常直观,从start或end开始,不断加入所有可到达的单词,直到最终到达另一端.本质上广度优先遍历图. 需要注意的是,拓展下一个单词时不能对字典进行枚举,因为字典里的单词太多.幸好单词本身都不长,所以直接枚举单词所有可能的变形,看看在dict中出现没有. 当然还不止这些,上面的做法仍然会超时,需要添加剪枝策略: 1. 如果某个单词在以前遍历过了,那么以后都不用再考…

题目: Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that: Only one letter can be changed at a time Each intermediate word must exist in the word list…

Word Ladder II Total Accepted: 11755 Total Submissions: 102776My Submissions Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each i…

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that: Only one letter can be changed at a time. Each transformed word must exist in the word li…

Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord to endWord, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For exam…

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, Given: start = "hi…

其他LeetCode题目欢迎访问:LeetCode结题报告索引 LeetCode:Word Ladder Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: Only one letter can be changed at a time Each intermediate word…

Word Ladder Total Accepted: 24823 Total Submissions: 135014My Submissions Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: Only one letter can be changed at a time Ea…

  Word Ladder II Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, G…

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, Given:start…

Problem Link: http://oj.leetcode.com/problems/word-ladder-ii/ Basically, this problem is same to Word Ladder I, which uses a double-direction BFS. However, the difference is that we need to keep track of all paths during the double-direction BFS in o…

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, Given:start…

原题地址 BFS Word Ladder II的简化版(参见这篇文章) 由于只需要计算步数,所以简单许多. 代码: int ladderLength(string start, string end, unordered_set<string> &dict) { if (start == end) ; unordered_set<string> old; queue<string> layer; ; layer.push(start); while (!laye…