图遍历算法的应用（包括输出长度为l的路径最短最长路径）

Posted 2020-11-22 kirosola

判断从顶点u到v是否有路径

void ExistPath(AdjGraph* G, int u, int v, bool& has)
{
	int w;
	ArcNode* p;
	visit[u] = 1;
	if (u == v)
	{
		has = true;
		return;
	}
	p = G->adjlist[u].firstarc;
	while (p != NULL)
	{
		w = p->adjvex;
		if (visit[w] == 0)
			ExistPath(G, w, v, has);
		p = p->nextarc;
	}
}

　　输出u到v的一条简单路径

void FindPath(AdjGraph* G, int u, int v, int path[], int d)
{
	path[++d] = u;
	visit[u] = 1;
	if (u == v)
	{
		for (int i = 0; i <= d; i++)
			printf("%d ", path[i]);
		printf("
");
		return;
	}
	ArcNode* p = G->adjlist[u].firstarc;
	while (p != NULL)
	{
		if (visit[p->adjvex] == 0)
			FindPath(G, p->adjvex, v, path, d);
		p = p->nextarc;
	}
}

　　输出u到v的所有简单路径，回溯的深度优先搜索算法

void FindAllPath(AdjGraph* G, int u, int v, int path[], int d)
{
	path[++d] = u;
	visit[u] = 1;
	if (u == v)
	{
		for (int i = 0; i <= d; i++)
			printf("%2d", path[i]);
		printf("
");
	}
	ArcNode* p = G->adjlist[u].firstarc;
	while (p!=NULL)
	{
		if (visit[p->adjvex] == 0)
			FindAllPath(G, p->adjvex, v, path, d);
		p = p->nextarc;
	}
	visit[u] = 0;
}

　　输出u到v长度为l的路径

void FindlenPath(AdjGraph* G, int u, int v, int l, int path[], int d)
{
	path[++d] = u;
	visit[u] = 1;
	if (u == v && d == l)
	{
		for (int i = 0; i <= d; i++)
			printf("%2d", path[i]);
		printf("
");
	}
	ArcNode* p = G->adjlist[u].firstarc;
	while (p != NULL)
	{
		if (visit[p->adjvex] == 0)
			FindlenPath(G, p->adjvex, v, l, path, d);
		p = p->nextarc;
	}
	visit[u] = 0;
} 

　　输出u到v的最短路径

typedef struct {
	int data;
	int parent;
}Queue;
void ShortPath(AdjGraph* G, int u, int v)
{
	int w;
	ArcNode* p;
	Queue qu[MAXV];
	int front = -1, rear = -1;
	int visit[MAXV] = { 0 };
	rear++;
	qu[rear].data = u;
	qu[rear].parent = -1;
	visit[u] = 1;
	while (front != rear)
	{
		front++;
		w = qu[front].data;
		if (w == v)
		{
			int i = front;
			while (qu[i].parent != -1)
			{
				printf("%2d", qu[i].data);
				i = qu[i].parent;
			
			}
			printf("%2d
", qu[i].data);
			return;
		}
		p = G->adjlist[w].firstarc;
		while (p != NULL)
		{
			if (visit[p->adjvex] == 0)
			{
				rear++;
				qu[rear].data = p->adjvex;
				qu[rear].parent = front;
				visit[p->adjvex] = 1;
			}
			p = p->nextarc;
		}
	}
}

　　求距离u最短的一个顶点

int Maxdist(AdjGraph* G, int v)
{
	ArcNode* p;
	int qu[MAXV];
	int rear = 0, front = 0;
	int visit[MAXV] = { 0 };
	int i, j, k;
	qu[++rear] = v;
	visit[v] = 1;
	while (rear != front)
	{
		front = (front + 1) % MAXV;
		k = qu[front];
		p = G->adjlist[k].firstarc;
		while (p != NULL)
		{
			if (visit[p->adjvex] == 0)
			{
				rear = (rear + 1) % MAXV;
				qu[rear] = p->adjvex;
				visit[p->adjvex] = 1;
			}
			p = p->nextarc;
		}
	}
	return k;
}

　　输出经过k的所有简单路径

void DFSPath(AdjGraph* G, int u, int v, int path[], int d)
{
	int w, i;
	visit[u] = 1;
	path[++d] = u;
	ArcNode* p = G->adjlist[u].firstarc;
	while (p != NULL)
	{
		w = p->adjvex;
		if (w == v && d > 1)
		{
			printf(" ");
			for (i = 0; i <= d; i++)
				printf("%d ", path[i]);
			printf("%d
",v);
		}
		if (visit[w] == 0)
			DFSPath(G, w, v, path, d);
		p = p->nextarc;
	}
	visit[u] = 0;
}
void FindCirclePath(AdjGraph* G, int k)
{
	int path[MAXV];
	DFSPath(G, k, k, path, -1);
}