基于矩阵实现的广度优先搜索
1.广度优先算法
广度优先算法是指，从指定结点，沿着图的宽度遍历所有节点。在Wikipedia上的定义如下：Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. It starts at the tree root (or some arbitrary node of a graph, sometimes referred to as a 'search key') and explores the neighbor nodes first, before moving to the next level neighbors.
2.传统算法
2.1算法思想
广度优先搜索类似于树的层次遍历。其算法思路如下：- 从图中的某个顶点V出发，访问之；并将其访问标志置为已被访问，即visited[i]=1；
- 依次访问顶点V的各个未被访问过的邻接 点，将V的全部邻接点都访问到；
- 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，并使“先被访问的顶 点的邻接点”先于“后被访问的顶点的邻接点”被访问，直到图中所有已被访问过的顶 点的邻接点都被访问到。
依此类推，直到图中所有顶点都被访问完为止。
2.2算法实现
下面是Java实现的无向图的广度优先搜索：
public class BreadthFirstPaths {
private static final int INFINITY = Integer.MAX_VALUE;
private boolean[] marked; // marked[v] = is there an s-v path
private int[] edgeTo; // edgeTo[v] = previous edge on shortest s-v path
private int[] distTo; // distTo[v] = number of edges shortest s-v path
/**
* Computes the shortest path between the source vertex <tt>s</tt>
* and every other vertex in the graph <tt>G</tt>.
* @param G the graph
* @param s the source vertex
*/
public BreadthFirstPaths(Graph G, int s) {
marked = new boolean[G.V()];
distTo = new int[G.V()];
edgeTo = new int[G.V()];
bfs(G, s);
assert check(G, s);
}
/**
* Computes the shortest path between any one of the source vertices in <tt>sources</tt>
* and every other vertex in graph <tt>G</tt>.
* @param G the graph
* @param sources the source vertices
*/
public BreadthFirstPaths(Graph G, Iterable<Integer> sources) {
marked = new boolean[G.V()];
distTo = new int[G.V()];
edgeTo = new int[G.V()];
for (int v = 0; v < G.V(); v++)
distTo[v] = INFINITY;
bfs(G, sources);
}
// breadth-first search from a single source
private void bfs(Graph G, int s) {
Queue<Integer> q = new Queue<Integer>();
for (int v = 0; v < G.V(); v++)
distTo[v] = INFINITY;
distTo[s] = 0;
marked[s] = true;
q.enqueue(s);
while (!q.isEmpty()) {
int v = q.dequeue();
for (int w : G.adj(v)) {
if (!marked[w]) {
edgeTo[w] = v;
distTo[w] = distTo[v] + 1;
marked[w] = true;
q.enqueue(w);
}
}
}
}
// breadth-first search from multiple sources
private void bfs(Graph G, Iterable<Integer> sources) {
Queue<Integer> q = new Queue<Integer>();
for (int s : sources) {
marked[s] = true;
distTo[s] = 0;
q.enqueue(s);
}
while (!q.isEmpty()) {
int v = q.dequeue();
for (int w : G.adj(v)) {
if (!marked[w]) {
edgeTo[w] = v;
distTo[w] = distTo[v] + 1;
marked[w] = true;
q.enqueue(w);
}
}
}
}
/**
* Is there a path between the source vertex <tt>s</tt> (or sources) and vertex <tt>v</tt>?
* @param v the vertex
* @return <tt>true</tt> if there is a path, and <tt>false</tt> otherwise
*/
public boolean hasPathTo(int v) {
return marked[v];
}
/**
* Returns the number of edges in a shortest path between the source vertex <tt>s</tt>
* (or sources) and vertex <tt>v</tt>?
* @param v the vertex
* @return the number of edges in a shortest path
*/
public int distTo(int v) {
return distTo[v];
}
/**
* Returns a shortest path between the source vertex <tt>s</tt> (or sources)
* and <tt>v</tt>, or <tt>null</tt> if no such path.
* @param v the vertex
* @return the sequence of vertices on a shortest path, as an Iterable
*/
public Iterable<Integer> pathTo(int v) {
if (!hasPathTo(v)) return null;
Stack<Integer> path = new Stack<Integer>();
int x;
for (x = v; distTo[x] != 0; x = edgeTo[x])
path.push(x);
path.push(x);
return path;
}
// check optimality conditions for single source
private boolean check(Graph G, int s) {
// check that the distance of s = 0
if (distTo[s] != 0) {
StdOut.println("distance of source " + s + " to itself = " + distTo[s]);
return false;
}
// check that for each edge v-w dist[w] <= dist[v] + 1
// provided v is reachable from s
for (int v = 0; v < G.V(); v++) {
for (int w : G.adj(v)) {
if (hasPathTo(v) != hasPathTo(w)) {
StdOut.println("edge " + v + "-" + w);
StdOut.println("hasPathTo(" + v + ") = " + hasPathTo(v));
StdOut.println("hasPathTo(" + w + ") = " + hasPathTo(w));
return false;
}
if (hasPathTo(v) && (distTo[w] > distTo[v] + 1)) {
StdOut.println("edge " + v + "-" + w);
StdOut.println("distTo[" + v + "] = " + distTo[v]);
StdOut.println("distTo[" + w + "] = " + distTo[w]);
return false;
}
}
}
// check that v = edgeTo[w] satisfies distTo[w] + distTo[v] + 1
// provided v is reachable from s
for (int w = 0; w < G.V(); w++) {
if (!hasPathTo(w) || w == s) continue;
int v = edgeTo[w];
if (distTo[w] != distTo[v] + 1) {
StdOut.println("shortest path edge " + v + "-" + w);
StdOut.println("distTo[" + v + "] = " + distTo[v]);
StdOut.println("distTo[" + w + "] = " + distTo[w]);
return false;
}
}
return true;
}
/**
* Unit tests the <tt>BreadthFirstPaths</tt> data type.
*/
public static void main(String[] args) {
In in = new In(args[0]);
Graph G = new Graph(in);
// StdOut.println(G);
int s = Integer.parseInt(args[1]);
BreadthFirstPaths bfs = new BreadthFirstPaths(G, s);
for (int v = 0; v < G.V(); v++) {
if (bfs.hasPathTo(v)) {
StdOut.printf("%d to %d (%d): ", s, v, bfs.distTo(v));
for (int x : bfs.pathTo(v)) {
if (x == s) StdOut.print(x);
else StdOut.print("-" + x);
}
StdOut.println();
}
else {
StdOut.printf("%d to %d (-): not connected\n", s, v);
}
}
}
}
3.基于矩阵的算法
3.1算法推导
邻接矩阵的定义形式：结点i和结点j邻接，则邻接矩阵A(i，j) =1；邻接矩阵对角线元素值为1，即A(i，i)=1；结点i和结点j不邻接，则邻接矩阵A(i，j)=0，可表示为：
表示y的第i行，包含结点i的邻结点，即y的第i行中非零值，是结点i经过一条路径连接的顶点。以此类推，经过若干次相乘，即经过若干次搜索，可以得到BFS搜索结果。用公式可以表示为：
其中BFS向量每经过一次乘运算后，新添加的非零值即为搜索结点。
另外，不难发现如果用
与对角矩阵X进行若干次相乘后，即可得到以图中每个结点为起点，进行BFS搜索后构成的结果矩阵。
3.2算法实现
用Python实现的代码：import numpy as np
from numpy.random import rand
def init(dimension, startVertex = 0):
mat = np.ones((dimension, dimension), dtype = np.int32)
mat[(rand(dimension ** 2 / 2) * dimension).astype(int), (rand(dimension ** 2 / 2) * dimension).astype(int)] = 0
for i in range(dimension):
mat[i, i] = 1
return mat
def bfsSingal(adjacencyMat, startVertex):
def nonZeroNum(vector):
return len(np.nonzero(vector)[0])
adjacencyMat = adjacencyMat.T
n = np.shape(adjacencyMat)[0]
current = np.zeros((n, 1), dtype = np.int32)
current[startVertex, 0] = 1
i = 0
stop = False
print("The %sth search: %s" % (i, current.T))
while(not stop):
i += 1
next = np.dot(adjacencyMat, current)
stop = nonZeroNum(current) == nonZeroNum(next)
if(not stop):
current = next
print("The %sth search: %s" % (i, current.T))
if __name__ == "__main__":
adjacencyMat = init(7, 3)
adjacencyMat1 = np.array([[1, 1, 0, 1, 0, 0, 0],
[0, 1, 0, 0, 1, 0, 1],
[0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 0],
[0, 0, 1, 0, 0, 1, 0],
[0, 0, 1, 1, 1, 0, 1]])
adjacencyMat2 = np.array([[1, 1, 0, 1, 0, 0, 0],
[0, 1, 0, 0, 1, 0, 1],
[0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 0],
[0, 0, 1, 0, 0, 1, 1],
[0, 0, 1, 1, 1, 0, 1]])
bfsSingal(adjacencyMat1, 3)