最短路径算法----Dijkatra

题目描述：对于上图中某个点i，求其他所有点和它的最近距离

适用于：没有负数边的情况；有向图、无向图都可以使用该算法

数据结构：可以使用二维矩阵代表图、也可以使用数据结构描述graph

算法：

1. 用到的数据集合：s（点k到点i最近的距离）, u（不在s中的其他点到i的距离）
2. 算法描述：
   1. 取u中距离最近的点j，加入s
   2. 对于和j相邻的点，更新u中的距离

def dijkstra(graph, start_node):# graph:n*n matrix# find min distance from start_nodes = {start_node:0}          # the minimum distance between node i and vu = {}                      # the distance between node i and v# init u: O(Vertex)for node, weight in enumerate(graph[start_node]):if node == start_node:continueif weight > 0:u[node] = weightelse:u[node] = 0xFFFFFF      # max distancewhile u:# 1. from u get min distance node# 2. update s# 3. update u using min distance node# O(Edge+Vertex^2) if using graph data structure# O(Vertex^2) is using matrixmin_dist_node = min(u, key=u.get)dist = u[min_dist_node]s[min_dist_node] = distdel u[min_dist_node]for node, weight in enumerate(graph[min_dist_node]):if node in u and weight > 0:u[node] = min(u[node], dist + weight)return s

使用：

graph = [[0, 7, 9,  0,  0, 14],         [7, 0, 10, 15, 0, 0],         [9, 10, 0, 11, 0, 2],         [0, 15, 11, 0, 6, 0],         [0, 0,  0,  6, 0, 9],         [14, 0, 2, 0,  9, 0]         ]print dijkstra(graph, 0)

性能分析：
如果使用图的数据结构，每条边只遍历一次，根据min获取s中最小边的方式，有不同的答案：

1.如果使用堆结构维护u的话：更新O(log(Vertex))，找最小节点O(1)。遍历n个节点，需要O(Vertex*log(Vertex))；更新边需要更新u，需要O(Edge*log(Vertex))。最终复杂度为O(Vertex*log(Vertex)+Edge*log(Vertex))

2. 如果使用map维护u的话：更新O(1)，找最小节点O(Vertex)。遍历n个节点，需要O(Vertex^2)；更新边只要O(Edge)。最终复杂度为O(Vertex^2 + Edge)

如果使用邻接矩阵，O(Vetex^2)

附录：带上最短路径pre_node_map

def dijkstra(graph, start_node):# graph:n*n matrix# find min distance from start_nodepre_node_map = {start_node: -1}s = {start_node:0}          # the minimum distance between node i and vu = {}                      # the distance between node i and v# init u: O(Vertex)for node, weight in enumerate(graph[start_node]):if node == start_node:continueif weight > 0:u[node] = weightpre_node_map[node] = start_nodeelse:u[node] = 0xFFFFFF      # max distancewhile u:# 1. from u get min distance node# 2. update s# 3. update u using min distance node# O(Edge) if using graph data structure# O(Vertex^2) is using matrixmin_dist_node = min(u, key=u.get)dist = u[min_dist_node]s[min_dist_node] = distdel u[min_dist_node]for node, weight in enumerate(graph[min_dist_node]):if node in u and weight > 0:if dist + weight < u[node]:u[node] = dist + weightpre_node_map[node] = min_dist_nodeprint "pre_node_map", pre_node_mapreturn s