图通常有两种表示方法: 邻接矩阵 和 邻接表

对于稀疏的图,邻接表表示能够极大地节省空间。

以下是图的数据结构的主要部分:

struct Vertex{

ElementType element; //节点的名字

Edge *next;   //所包含的边组成的单链表的头指针

};

struct Edge{

int adj;  //节点的标号(0-number of nodes)

Edge *next;

};

注意,实际应用中,节点都有名字,而不是数字,所以我们需要提供从名字到标号的映射。

最简单的方法就是 Hash(散列表),或者二叉查找树之类的能够快速提供查询的数据结构。

本文的处理较为简单: 节点的名字是 'a' ---- 'z', 所以映射通过简单的运算即可实现:   node id = node name - 'a'.

// copyright @ L.J.SHOU Jan.13, 2014

#include "graph.h"
#include <ctime>
#include <iostream>
#include <stack>
#include <queue>
using namespace std; typedef char ElementType;
typedef Vertex* Graph; enum Color{WHITE, GRAY, BLACK}; struct Edge
{
int vertex;
int weight;
Edge *next;
}; struct Vertex
{
Color color;
ElementType element;
int start, finish;
Edge *next;//head of adjacent list
}; Graph Create(Graph graph, int n)
{
graph = new Vertex[n];
for(int i=0; i<n; ++i)
{
graph[i].color = WHITE;
graph[i].element = i + 'a';
graph[i].start = 0;
graph[i].finish = 0;
graph[i].next = NULL;
}
return graph;
} // Reset Graph
void Clear(Graph graph, int n)
{
for(int i=0; i<n; ++i)
{
graph[i].color = WHITE;
graph[i].start = 0;
graph[i].finish = 0;
}
} Graph DeleteGraph(Graph graph, int n)
{
for(int i=0; i<n; ++i)
{
Edge* head(graph[i].next), *next(NULL);
while(head)
{
next = head->next;
delete head;
head = next;
}
}
delete [] graph;
return NULL;
} // return the outdegree of vertex i
int OutDegree(Graph g, int i)
{
int num(0); Edge* link(g[i].next);
while(link)
{
link = link->next;
++ num;
}
return num;
} // test whether edge(i, j) exists
bool Exist(Graph g, int i, int j)
{
Edge *link(g[i].next); while(link && link->vertex != j)
link = link->next; if(link == NULL)
return false;
else
return true;
} bool InsertEdge(Graph g, int i, int j)
{
if(Exist(g, i, j)){
cout << "edge (" << i << "," << j << ") already existed" << endl;
return false;
} Edge *edge(NULL);
edge = new struct Edge;
edge->vertex = j;
edge->next = g[i].next;
g[i].next = edge; return true;
} bool DeleteEdge(Graph g, int i, int j)
{
if(!Exist(g, i, j)){
cout << "edge (" << i << "," << j << ") doesn't exist" << endl;
return false;
} Edge *cur(g[i].next), *pre(cur); while(cur && cur->vertex != j)
{
pre = cur;
cur = cur->next;
} if(pre == NULL)
{ // delete head edge
g[i].next = cur->next;
delete cur;
}
else
{
pre->next = cur->next;
delete cur;
}
return true;
} // print adjacent list
void OutPut(Graph g, int n)
{
Edge *edge(NULL);
for(int i=0; i<n; ++i)
{
cout << g[i].element << "->";
edge = g[i].next;
while(edge)
{
cout << g[edge->vertex].element << "->";
edge = edge->next;
}
cout << "NULL" << endl;
}
} void DFS(Graph graph, int n)
{
cout << "DFS: " << endl;;
Clear(graph, n);
for(int i=0; i<n; ++i)
{
if(graph[i].color == WHITE)
DFSVisit(graph, i);
}
cout << endl; cout << "DFS_stack: " << endl;
Clear(graph, n);
for(int i=0; i<n; ++i)
{
if(graph[i].color == WHITE)
DFSVisitStack(graph, i);
}
cout << endl;
} // recursive DFS
void DFSVisit(Graph graph, int i)
{
static int time(0);
Edge *link(graph[i].next); cout << graph[i].element << " ";
graph[i].color = GRAY;
graph[i].start = ++time; while(link)
{
if(graph[link->vertex].color == WHITE)
DFSVisit(graph, link->vertex);
link = link->next;
} graph[i].finish = ++time;
graph[i].color = BLACK;
} // non-recursive DFS
void DFSVisitStack(Graph g, int i)
{
static int time(0);
struct Edge* edge;
int vertex;
stack<int> s; //visit vertex i
cout << g[i].element << " ";
g[i].color = GRAY;
g[i].start = ++time;
s.push(i); while(!s.empty())
{
vertex = s.top();
edge = g[vertex].next;
while(edge)
{
if(g[edge->vertex].color == WHITE)
{
s.push(edge->vertex);
cout << g[edge->vertex].element << " ";
g[edge->vertex].start = ++time;
g[edge->vertex].color = GRAY;
break;
}
edge = edge->next;
}
//vertex's neigbours have been visited
if(edge == NULL){
s.pop();
g[vertex].color = BLACK;
g[vertex].finish = ++time;
}
}
} /////////////////////////////////////////////////////////////
// search all vertices that can be rearched from Source s ///
// compute the distances from source s /// ///////////////
/////////////////////////////////////////////////////////////
void BFS(Graph g, int n, int s)
{
queue<int> q;
Edge *edge(NULL);
int vertex; //visit source vertex
Clear(g, n);
cout << "BFS: " << endl;;
cout << g[s].element << " ";
g[s].color = GRAY;
q.push(s); while(!q.empty())
{
//dequeue
vertex = q.front();
q.pop(); //all the adjacent vertices
edge = g[vertex].next;
while(edge)
{
if(g[edge->vertex].color == WHITE){
g[edge->vertex].color = GRAY;
cout << g[edge->vertex].element << " ";
//enqueue
q.push(edge->vertex);
}
edge = edge->next;
}
g[vertex].color = BLACK;
}//end of while cout << endl;
} int main()
{
Graph graph;
int num_vertices = 8; graph = Create(graph, num_vertices); InsertEdge(graph,0,1);
InsertEdge(graph,1,2);
InsertEdge(graph,2,3);
InsertEdge(graph,3,2);
InsertEdge(graph,4,0);
InsertEdge(graph,1,5);
InsertEdge(graph,2,6);
InsertEdge(graph,3,7);
InsertEdge(graph,1,4);
InsertEdge(graph,4,5);
InsertEdge(graph,5,6);
InsertEdge(graph,6,7);
InsertEdge(graph,7,7);
InsertEdge(graph,6,5); OutPut(graph, num_vertices);
DFS(graph, num_vertices);
BFS(graph, num_vertices, 0); graph = DeleteGraph(graph, num_vertices); return 0;
}

数据结构 《2》----基于邻接表表示的图的实现 DFS(递归和非递归), BFS的更多相关文章

  1. 图的基本操作(基于邻接表):图的构造,深搜(DFS),广搜(BFS)

    #include <iostream> #include <string> #include <queue> using namespace std; //表结点 ...

  2. SDUT 2142 数据结构实验之图论二:基于邻接表的广度优先搜索遍历

    数据结构实验之图论二:基于邻接表的广度优先搜索遍历 Time Limit: 1000MS Memory Limit: 65536KB Submit Statistic Problem Descript ...

  3. 数据结构C++使用邻接表实现图

    定义邻接表存储的图类.[实验要求] (1)创建一个邻接表存储的图:(2)返回图中指定边的权值:(3)插入操作:向图中插入一个顶点,插入一条边:(4)删除操作:从图中删除一个顶点,删除一条边:(5)图的 ...

  4. 无向图的 DFS 和 BFS实现 (以邻接表存储的图)

    #include <iostream> #include <queue> using namespace std; #define MaxVertexNum 10 typede ...

  5. 数据结构二叉树的递归与非递归遍历之java,javascript,php实现可编译(1)java

    前一段时间,学习数据结构的各种算法,概念不难理解,只是被C++的指针给弄的犯糊涂,于是用java,web,javascript,分别去实现数据结构的各种算法. 二叉树的遍历,本分享只是以二叉树中的先序 ...

  6. 数据结构作业——图的存储及遍历(邻接矩阵、邻接表+DFS递归、非递归+BFS)

    邻接矩阵存图 /* * @Author: WZY * @School: HPU * @Date: 2018-11-02 18:35:27 * @Last Modified by: WZY * @Las ...

  7. SDUT2142数据结构实验之图论二:基于邻接表的广度优先搜索遍历

    http://acm.sdut.edu.cn/sdutoj/showproblem.php?pid=2142&cid=1186 题目描述 给定一个无向连通图,顶点编号从0到n-1,用广度优先搜 ...

  8. Invitation Cards(邻接表+逆向建图+SPFA)

    Time Limit: 8000MS   Memory Limit: 262144K Total Submissions: 17538   Accepted: 5721 Description In ...

  9. 基于邻接表的长度为k的简单路径的求解

    描述 一个连通图采用邻接表作为存储结构.设计一个算法,判断无向图中任意给定的两点是否存在一条长度为k的简单路径. 输入 多组数据,每组m+3数据行.第一行有两个数字n,m和k,代表有n个顶点,m条边和 ...

随机推荐

  1. robotframework笔记24

    侦听器接口 机器人框架有一个侦听器接口,可以用于接收 对测试执行通知. 示例用法包括 外部测试监控,测试失败时发送邮件消息, 与其他系统进行通信. 侦听器API版本3也使得 它可以修改测试在测试执行和 ...

  2. 为设计师准备的 20 个新的免费 PSD 模板

    设计师们每天的工作经常要用到各种模板,本文推荐 20 个新的免费 PSD 模板,可节省设计时间. Blueprint Notice PSD Wooden Drawer Icon Minimal Clo ...

  3. Shell基础:变量类型 & 运算符

    Shell变量 Shell支持三种类型的变量 用户自定义变量:用户自定义的变量,变量名以英文字母或下划线开头,区分大小写. 位置变量:根据位置传递参数给脚本的变量,默认支持9个位置变量 $1,$2,$ ...

  4. hdu-----(1507)Uncle Tom's Inherited Land*(二分匹配)

    Uncle Tom's Inherited Land* Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (J ...

  5. 离线网页制作器(beta1.0)

    package hhuarongdao; /* *使用方法: 先选择保存路径,然后输入相应的网址, *然后会得到那个网页的离线版的 内容 * */ import java.awt.BorderLayo ...

  6. 你不知道的JavaScript--大白话讲解Promise

    转载:http://blog.csdn.net/i10630226/article/details/50867792 一.Promise小试 复杂的概念先不讲,我们先简单粗暴地把Promise用一下, ...

  7. Eclipse如何设置代码提示功能

    Windows→Preference→XML→XML Files→Editor→Content Assist→Auto Activation→Prompt when these characters ...

  8. js中获得当前时间是年份和月份

    js中获得当前时间是年份和月份,形如:201208       //获取完整的日期 var date=new Date; var year=date.getFullYear();  var month ...

  9. 好用的json-path

    $.store.book[?(@.price < 10)].title Here is a complete overview and a side by side comparison of ...

  10. ASP.NET MVC的Ajax.ActionLink 的HttpMethod="Get" 一个重复请求的BUG

    这段时间使用BootStrap+Asp.net Mvc5开发项目,Ajax.ActionLink遇到一个重复提交的BUG,代码如下: @model IList<WFModel.WF_Temp&g ...