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Data structure adjacency multiple table (C language implementation)

2020-11-10 10:51:52 3fc4y

I'm learning data structure recently , Yesterday I realized the adjacency multiple table , There was a little problem before writing , Originally wanted to find a big man to write the program reference , But there was no satisfactory , So I can only write by myself . Little brother wrote this program, the whole process only refers to the textbook on the adjacency multiple table some simple text description , As for the code section , It's all written by my little brother alone , There is no reference , The writing is redundant , So if some big guy thinks that my brother's writing is too clumsy , Please keep your mouth on your mind , First time I posted an article on the Internet .

Compared with adjacency list, adjacency multiple table saves space greatly ( Half ), In adjacency multiple table, when defining the structure of adjacent vertices, it does not represent a vertex , It's a side by node_1 To node_2 One side of , And there are two pointer fields, one of which belongs to node_1 A pointer field belongs to node_2, They two don't interfere with , It's like two adjacent highways , Put on a road node_1 Motorcade , Put on another road node_2 Motorcade ( First of all, you should create a sense of adjacency to multiple tables ), Therefore, an edge node can be used by two vertices
This is the structure that defines the edge nodes
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This is the structure that defines the vertex nodes , among vex Record top points ,edge Record edges ,head The information of each vertex is recorded in the structure , And the pointer field of the linked list
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This is a picture on the Internet adjacent to multiple tables , You can understand
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Then I'll talk about the key points and difficulties of this program !!!!!
Look at the example above , Let's find any side , Just v2 To v5 This side , That is to say (4,1) This side , What we need to do now is mount the side to v2 Nodes and v5 Under the node , Let's be simple , Let's mount this side in v5 Next ( Mount on the with node_1 Under vertex nodes with the same value ), First of all, judge v5 Whether there are mount nodes under , Judge nothing , Then mount the node directly in v5 Vertex node of next in , This is the success of the mount .
Then mount it on v2 Next ( Mount on the with node_2 Under vertex nodes with the same value ), equally , First judge v2 Is there a mount node , There is a judgment , use 1(node_2) And v2 The first node of (0,1) Compare ( First of all, determine the attached edge node's node_1 and node_2 Which vertex to mount v2). Found the mount of node_2 And the ones to be mounted node_2 identical , And then compare what you want to mount node_1(4) Larger than the one that has been mounted node_1(0), Get into node_2_pointer, And again (2,1) Comparison found mounted node_2 And the ones to be mounted node_2 identical , And then compare what you want to mount node_1(4) Larger than the one that has been mounted node_1(2), Get into node_2_pointer, Find out node_2_pointer by NULL Will (4,1) Edge node , Mount this pointer field . Those who have doubts about the above operation can take a look at the contents in brackets below .
(------------------------------------------------------------------------------------------------------------------
Because it is to be inserted with node_2 Under the same vertex node , So we need to use what we want to mount node_2 Make the following judgments :
Case one : To mount the edge node of node_2 And the ones that have been mounted node_1 identical , And it's better to mount node_1 And the ones that have been mounted node_2, If small , Then the edge node is inserted into the first (next), Otherwise , The node pointer enters the mounted node node_1_pointer( Why not enter node_2_pointer Well ? Because the above has been compared to mount node_2 With the ones that have been mounted node_1 The value of is the same , So to enter node_1_pointer Pointer to the domain ), And then continue with the same operation ( Compare -> Judge the size -> Insert a node or enter a pointer field ).
The second case : To mount the edge node node_2 With the ones that have been mounted node_2 identical , And it's better to mount node_1 And the ones that have been mounted node_1, If small , Then the edge node is inserted into the first (next), Otherwise , The node pointer enters the mounted node node_2_pointer, Why enter the pointer field , The reason as above . And then continue with the same operation .
)--------------------------------------------------------------------------------------------------------------------








in general , There are three cases of insertion ,
Case one : No mount on the vertex head , Or judge that the adjacent node of the first edge node that has been mounted is larger than the adjacent node of the edge node to be mounted , Then the edge node is inserted into the vertex node next
The second case : Determine that it needs to be inserted in the middle of the mount list of vertex nodes
The third case : The adjacent nodes of the edge nodes to be mounted are larger than those of all the mounted edge nodes
( Because it's too complicated , So language is very difficult to describe , You should look the same ^ _ ^)
Now please take a look at the code




#include<stdio.h>
// Adjacency multiple tables 
#define MAX 20


struct Enode
{
   
   
	int node_1;
	int node_2;
	struct Enode* node_1_pointer;
	struct Enode* node_2_pointer;
};

struct AMLGraph
{
   
   
	int edge;
	int vex;
	struct
	{
   
   
		char head_ele;
		struct Enode* next;
	}head[MAX];
};

void show(struct AMLGraph* map);
void creat(struct AMLGraph* map);
void insert(struct AMLGraph* map, char node_1, char node_2);


void creat(struct AMLGraph* map)
{
   
   
	char node_1;
	char node_2;
	printf(" To build several nodes , A few sides :");
	scanf("%d%d", &map->vex, &map->edge);
	getchar();
	for (int i = 0; i < map->vex; i++)
	{
   
   
		map->head[i].next = NULL;
		map->head[i].head_ele = i + 'A';
	}
	printf(" Please input side :\n");
	for (int i = 0; i < map->edge; i++)
	{
   
   
		scanf("%c%c", &node_1, &node_2);
		getchar();
		insert(map, node_1, node_2);
	}


}
void insert(struct AMLGraph* map, char node_1, char node_2)
{
   
   
	struct Enode* now;
	struct Enode* q;
	struct Enode* before;
	struct Enode* p = malloc(sizeof(struct Enode));
	p->node_1 = node_1 - 'A';
	p->node_2 = node_2 - 'A';
	p->node_1_pointer = NULL;
	p->node_2_pointer = NULL;
	// Mount this node in node_1 and node_2 In the linked list 
	// Insert and node_1 The same list 
	now = map->head[node_1 - 'A'].next;
	before = now;
	while (now)
	{
   
   
		// Determine whether the node should be inserted in the chain header 
		if (node_1-'A' == map->head[node_1 - 'A'].next->node_1 )
		{
   
   
			if (node_2-'A' < map->head[node_1 - 'A'].next->node_2)
			{
   
   
				p->node_1_pointer = map->head[node_1 - 'A'].next;
				map->head[node_1 - 'A'].next = p;
				break;
			}
		}
		else
		{
   
   
			if (node_2 - 'A' < map->head[node_1 - 'A'].next->node_1)
			{
   
   
				p->node_1_pointer = map->head[node_1 - 'A'].next;
				map->head[node_1 - 'A'].next = p;
				break;
			}
		}
		
		//node_1 and now->node_1 equal 
		if (node_1 - 'A' == now->node_1)
		{
   
   
			if (node_2 - 'A' > now->node_2)
			{
   
   
				before = now;
				now = now->node_1_pointer;
			}
			else if (node_2 - 'A' < now->node_2)
			{
   
   
				if (before->node_1 == node_1 - 'A')
				{
   
   
					q = before->node_1_pointer;
					before->node_1_pointer = p;
					p->node_1_pointer = q;
				}
				else
				{
   
   
					q = before->node_2_pointer;
					before->node_2_pointer = p;
					p->node_1_pointer = q;
				}
				break;
			}
		}
		else//node_1 and now->node_2 equal 
		{
   
   
			if (node_2 - 'A' > now->node_1)
			{
   
   
				before = now;
				now = now->node_2_pointer;
			}
			else if (node_2 - 'A' < now->node_1)
			{
   
   
				if (before->node_2 == node_1 - 'A')
				{
   
   
					q = before->node_2_pointer;
					before->node_2_pointer = p;
					p->node_1_pointer = q;
				}
				else
				{
   
   
					q = before->node_1_pointer;
					before->node_1_pointer = p;
					p->node_1_pointer = q;
				}
				break;
			}
		}

	}

	if (!now)// There are two situations map->head[node_1 - 'A'].next Empty or to be inserted at the end of the list 
	{
   
   
		if (!before)
		{
   
   
			map->head[node_1 - 'A'].next = p;
		}
		else if (before->node_1 == node_1 - 'A')
		{
   
   
			before->node_1_pointer = p;
		}
		else if (before->node_2 == node_1 - 'A')
		{
   
   
			before->node_2_pointer = p;
		}
	}
	//--------------------------------------------------------------------------------------
	// Insert and node_2 The same list 
	now = map->head[node_2 - 'A'].next;
	before = now;
	while (now)
	{
   
   
		// Determine whether the node should be inserted in the chain header 
		if (node_2 - 'A' == map->head[node_2 - 'A'].next->node_1)
		{
   
   
			if (node_1 - 'A' < map->head[node_2 - 'A'].next->node_2)
			{
   
   
				p->node_2_pointer = map->head[node_2 - 'A'].next;
				map->head[node_2 - 'A'].next = p;
				break;
			}
		}
		else
		{
   
   
			if (node_1 - 'A' < map->head[node_2 - 'A'].next->node_1)
			{
   
   
				p->node_2_pointer = map->head[node_2 - 'A'].next;
				map->head[node_2 - 'A'].next = p;
				break;
			}
		}

		//node_2 and now->node_2 equal 
		if (node_2 - 'A' == now->node_2)
		{
   
   
			if (node_1 - 'A' > now->node_1)		
			{
   
   
				before = now;
				now = now->node_2_pointer;
			}
			else if (node_1 - 'A' < now->node_1)
			{
   
   
				
				if (before->node_1 == node_2 - 'A')
				{
   
   
					q = before->node_1_pointer;
					before->node_1_pointer = p;
					p->node_2_pointer = q;
				}
				else
				{
   
   
					q = before->node_2_pointer;
					before->node_2_pointer = p;
					p->node_2_pointer = q;
				}
				break;
			}
		}
		else//node_2 and now->node_1 equal 
		{
   
   
			if (node_1 - 'A' > now->node_2)
			{
   
   
				before = now;
				now = now->node_1_pointer;
			}
			else if (node_1 - 'A' < now->node_2)
			{
   
   
				if (before->node_1 == node_2-'A')
				{
   
   
					q = before->node_1_pointer;
					before->node_1_pointer = p;
					p->node_2_pointer = q;
				}
				else
				{
   
   
					q = before->node_2_pointer;
					before->node_2_pointer = p;
					p->node_2_pointer = q;
				}
				break;
			}
		}
	}
	if (!now)// There are two situations map->head[node_2 - 'A'].next Empty or to be inserted at the end of the list 
	{
   
   
		if (!before)
		{
   
   
			map->head[node_2 - 'A'].next = p;
		}
		else if (before->node_1 == node_2 - 'A')
		{
   
   
			before->node_1_pointer = p;
		}
		else if (before->node_2 == node_2 - 'A')
		{
   
   
			before->node_2_pointer = p;
		}
	}
}

void show(struct AMLGraph* map)
{
   
   
    int head_ele;
	struct Enode* p;
	for (int i = 0; i < map->vex; i++)
	{
   
   
		p = map->head[i].next;
		head_ele = map->head[i].head_ele - 'A';
		printf(" And %c The connected nodes are :", head_ele+'A');
		while (p)
		{
   
   

			printf("%c ",( head_ele == p->node_1 ? (p->node_2 + 'A') :( p->node_1 + 'A')));//head_ele If it is equal to p->node_1 Just output another value 
			p = (head_ele == p->node_1 ?( p->node_1_pointer) : (p->node_2_pointer));//head_ele If it is equal to p->node_1 To get into p->node_1_pointer
		}
		printf("\n");
	}

}



int main()
{
   
   

	struct AMLGraph map;
	creat(&map);
	show(&map);
}





Look at two below vs2019 Run the results next

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I also found a more complex diagram in the book to verify , The result is correct
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If you don't understand, you can leave a message below , Make progress together ^ _ ^( by the way , If a friend runs the program and finds bug Leave a comment below , I can modify )

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