4_1邻接矩阵图

1. 图的邻接矩阵结构体定义

#include<stdio.h>
# define Maxvertices 20

typedef  struct Seqlist {
	int length;  //存储顶点总数
	char* data;  // 等价于 char data[Maxvertices]
}Seqlist;

typedef struct {

	Seqlist vertices;  //存放顶点的顺序表(顶点值和顶点的总数)
	int edge[Maxvertices][Maxvertices];//存放边的领接矩阵
	int numOfEdges;//存放边的条数
}AdjMGraph;

2. 连通图的bfs

//连通图的bfs
int bfs(AdjMGraph g) {

	int  queue[Maxvertices]; //存贮顶点的序号 ，也就是 顶点在数组中的下标。
	int front = 0;
	int rear = 0;

	int visit[Maxvertices] = { 0 };

	//设把0号顶点作为启动点。
	rear = (rear + 1) % Maxvertices;
	queue[rear] = 0;
	visit[0] = 1;

	while (rear != front)
	{
		front = (front + 1) % Maxvertices;
		int number = queue[front];

		printf("%c,", g.vertices.data[number]);

		for (int i = 0; i < g.vertices.length; i++)
		{
			if (g.edge[number][i] == 1 && visit[i] == 0)
			{
				rear = (rear + 1) % Maxvertices;
				queue[rear] = i;

				visit[i] = 1;

			}

		}
	}
	return 0;

}

3. 连通图的dfs

int dfs(AdjMGraph g) {

	int arr[Maxvertices];
	int top = -1;

	int visit[Maxvertices] = { 0 };


	visit[0] = 1;
	top++;
	arr[top] = 0;

	printf("%c", g.vertices.data[0]);

	while (top != -1)
	{
		int number = arr[top];
		top--;

		for (int i = 0; i < g.vertices.length; i++)
		{
			if (g.edge[number][i] == 1 && visit[i] == 0)
			{
				top++;
				arr[top] = number;

				top++;
				arr[top] = i;

				printf("%c", g.vertices.data[i]);

				visit[i] = 1;

				break;
			}
		}
	}
	return 0;

}

4. 一般图的bfs

//一般图的bfs

int bfs2(AdjMGraph g) {

	int queue[Maxvertices];
	int front = 0;
	int rear = 0;
	int visit[Maxvertices] = { 0 };

	for (int i = 0; i < g.vertices.length; i++)
	{
		if (visit[i] == 0)
		{
			rear = (rear + 1) % Maxvertices;
			queue[rear] = i;
			visit[i] = 1;
			while (rear != front) {

				front = (front + 1) % Maxvertices;
				int number = queue[front];


				printf("%c", g.vertices.data[number]);


				for (int i = 0; i < g.vertices.length; i++)
				{

					if (g.edge[number][i] == 1 && visit[i] == 0)
					{

						rear = (rear + 1) % Maxvertices;

						queue[rear] = i;

						visit[i] = 1;

					}
				}
			}
		}
	}






	return 0;

}

5. 一般图的dfs

//一般图的dfs
int dfs2(AdjMGraph g) {
	int arr[Maxvertices];
	int top = -1;

	int visit[Maxvertices] = { 0 };


	for (int i = 0; i < g.vertices.length; i++)
	{

		if (visit[i] == 0)
		{
			top++;
			arr[top] = i;
			visit[i] = 1;
			printf("%c", g.vertices.data[i]);

			while (top != -1) {

				int number = arr[top];
				top--;

				for (int j = 0; j < g.vertices.length; j++)
				{

					if (g.edge[number][j] == 1 && visit[j] == 0)
					{

						top++;
						arr[top] = number;

						top++;
						arr[top] = j;
						visit[j] = 1;

						printf("%c", g.vertices.data[j]);
						break;
					}

				}

			}

		}

	}

}

6. 1 判断图上任意点v点到达j点是否有路径可达

以v点作为启动点，通过深度遍历或者宽度遍历 如果遇到了k说明有可达的路径，否则没有

////
int v_j_lu_jing_dfs(AdjMGraph g, char v, char k) {
	int res = -1;

	int arr[Maxvertices];
	int top = -1;
	int visit[Maxvertices] = { 0 };

	//遍历 顶点数组，找到v点在数组中的下表，以加入栈中，作为启动点

	int index_v = -1;
	for (int i = 0; i < g.vertices.length; i++)
	{
		if (g.vertices.data[i] == v)
		{
			index_v = i;
			break;

		}

	}

	if (index_v == -1)
	{
		return -1;
	}

	top++;
	arr[top] = index_v;
	visit[index_v] = 1;

	while (top != -1)
	{
		int number = arr[top];
		top--;

		for (int i = 0; i < g.vertices.length; i++)
		{
			if (g.edge[number][i] == 1 && visit[i] == 0)
			{
				top++;
				arr[top] = number;

				top++;
				arr[top] = i;
				visit[i] = 1;

				if (g.vertices.data[i] == k)
				{
					return 1;
				}

			}
		}
	}
	return res;
}

int v_j_lu_jing_bfs(AdjMGraph g, char v, char k) {
	int res = -1;

	int queue[Maxvertices];
	int front = 0;
	int rear = 0;
	int visit[Maxvertices] = { 0 };
	
	int index_v = -1;

	for (int i = 0; i < g.vertices.length; i++)
	{
		if (g.vertices.data[i] == v)
		{
			index_v = i;
			break;
		}
	}
	if (index_v == -1)
	{
		return -1;
	}

	rear = (rear + 1) % Maxvertices;
	queue[rear] = index_v;

	visit[index_v] = 1;

	while (rear != front)
	{
		front = (front + 1) % Maxvertices;
		int number = queue[front];

		if (g.vertices.data[number] == k)
		{

			return 1;
		}

		for (int i = 0; i < g.vertices.length; i++)
		{
			if (g.edge[number][i] == 1 && visit[i] == 0)
			{

				rear = (rear + 1) % Maxvertices;
				queue[rear] = i;
				visit[i] = 1;

			}
		}

	}

	return res;


}

6.2 是否存在一个从v点到j点路径长度为k的路径

int v_j_lu_jing_changdu_k_bfs(AdjMGraph g, char v, char j, int k) {

	int res = 1;//保存最后要返回的结果 ，是否从在长度为k的路径，从v到j

	int queue[Maxvertices];
	int front = 0;
	int rear = 0;
	int visit[Maxvertices] = { 0 };

	//定义一个数组，用来存贮每个节点所在的层号。用顶点的序号作为存贮在数组的索引。通过顶点的序号，来取出顶点的层号
	int level_number[Maxvertices] = { 0 };

	int index_v = -1; //保存v点的下标 也就是v点的序号。
	for (int i = 0; i < g.vertices.length; i++)
	{

		if (g.vertices.data[i] == v)
		{
			index_v = i;
			break;
		}

	}

	if (index_v == -1)
	{
		return -1;
	}

	rear = (rear + 1) % Maxvertices;
	queue[rear] = index_v;
	visit[index_v] = 1;

	//设置启动顶点所在的层为第1层

	level_number[index_v] = 1;

	while (rear != front)
	{
		front = (front + 1) % Maxvertices;
		int number = queue[front];

		int cur_level_number = level_number[number];

		if (g.vertices.data[number] == j && k == cur_level_number - 1)
		{
			return 1;
		}

		for (int i = 0; i < g.vertices.length; i++)
		{

			if (g.edge[number][i] == 1 && visit[i] == 0)
			{

				rear = (rear + 1) % Maxvertices;
				queue[rear] = i;
				visit[i] = 1;

				level_number[i] = cur_level_number + 1;

			}

		}

	}

}

6.3 问无向图是不是一个连通图

实际上是问 极大连通分量是不是1，如果是的话就是连通图，如果不是就不是

6.4求无向图的连图分量个数

int lian_tong_fen_liang_ge_shu_bfs(AdjMGraph g) {
	int count = 0;

	int queue[Maxvertices];
	int front = 0;
	int rear = 0;
	int visit[Maxvertices] = { 0 };


	for (int i = 0; i < g.vertices.length; i++)
	{
		if (visit[i] == 0)
		{

			count++;
			rear = (rear + 1) % Maxvertices;
			queue[rear] = i;

			visit[i] = 1;

			while (rear != front)
			{
				front = (front + 1) % Maxvertices;
				int number = queue[front];

				for (int i = 0; i < g.vertices.length; i++)
				{
					if (g.edge[number][i] == 1 && visit[i] == 0)
					{

						rear = (rear + 1) % Maxvertices;
						queue[rear] = i;
						visit[i] = 1;

					}
				}

			}




		}
	}

	return count;



}

6.5 求无向图的连同分量，并打印各个连同分量的顶点。

int liang_tong_fen_liang_ge_shu_dayin_ding_dian_bfs(AdjMGraph g) {

	int count = 0;

	int queue[Maxvertices];
	int front = 0;
	int rear = 0;
	int visit[Maxvertices] = { 0 };


	for (int i = 0; i < g.vertices.length; i++)
	{
		if (visit[i] == 0)
		{

			count++;

			printf("第%d个连通分量:", count);

			rear = (rear + 1) % Maxvertices;
			queue[rear] = i;

			visit[i] = 1;

			while (rear != front)
			{
				front = (front + 1) % Maxvertices;
				int number = queue[front];

				printf("%c", g.vertices.data[number]);

				for (int i = 0; i < g.vertices.length; i++)
				{
					if (g.edge[number][i] == 1 && visit[i] == 0)
					{

						rear = (rear + 1) % Maxvertices;
						queue[rear] = i;
						visit[i] = 1;

					}
				}

			}




		}
	}

	return count;

}

6.6 求距离顶点v最远的顶点

以v点做启动点 进行宽度遍历，最后退出的点就是 距离v最远的点



char ju_v_zui_yuan_ju_li(AdjMGraph g, char v) {
	char res;

	int queue[Maxvertices];
	int front = 0;
	int rear = 0;

	int visit[Maxvertices] = { 0 };

	int index_v = -1;

	for (int i = 0; i < g.vertices.length; i++)
	{
		if (g.vertices.data[i] == v)
		{
			index_v = i;
			break;
		}
	}

	if (index_v == -1)
	{
		return -1;
	}

	rear = (rear + 1) % Maxvertices;
	queue[rear] = index_v;
	visit[index_v] = 1;

	while (rear != front)
	{
		front = (front + 1) % Maxvertices;
		int number = queue[front];

		res = g.vertices.data[number];

		for (int i = 0; i < g.vertices.length; i++)
		{
			if (g.edge[number][i] == 1 && visit[i] == 0)
			{
				rear = (rear + 1) % Maxvertices;
				queue[rear] = i;
				visit[i] = 1;

			}

		}
	}

	return res;




}

6.7判断一个图是否存在环

利用深度遍历 某个节点在未退站之前 ，又通过其他顶点的相邻顶点找到了该顶点

posted @   XjtuSeLwh2017  阅读(67)  评论(0)  编辑  收藏  举报
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