求出最短路径，要过程，用Dijkstra算法。。。

Posted 2023-03-21

参考技术A 从v1开始遍历
v2 = 2;
v3 = 5;
v2较小所以跳到v2
v3 = 4;

v4 = 6;
v5 = 8;
v3较小所以跳到v3
v4 = 5;
v6 = 7;
v4较小所以跳到v4
v6 = 6;
v7 = 9;
v6较小所以跳到v6
v7 = 8;

所以最后结果v1 -> v7最短路径为v1->v2->v3->v4->v6->v7,最短路径长度为8 参考技术B

# -*- coding: utf-8 -*-
"""
Spyder Editor

This is a temporary script file.
"""
#import collections
import heapq

#==============================================================================
# class Graph:
#     def __init__(self):
#         self.edges =
#         
#     def neighbors(self,id):
#         return self.edges[id]
#==============================================================================

#   原型
class GraphWithWeights():
    def __init__(self):
        #super.__init__(self)
        self.edges_weights =
    def cost(self, from_node, to_node):
        return self.edges_weights.get(from_node).get(to_node)
    def neighbors(self,node):
        return list(self.edges_weights.get(node).keys())

class PriorityQueue:
    def __init__(self):
        self.elements = []
#   检查优先序列是否为空
    def empty(self):
        return len(self.elements) == 0
#   依据优先权重添加一个成员    
    def put(self, item, priority):
        heapq.heappush(self.elements, (priority, item))
#   取出成员    
    def get(self):
        return heapq.heappop(self.elements)[1]

#   根据图设置值，顶点，边，权重，        
target_graph = GraphWithWeights()
target_graph.edges_weights =
    'V1':'V2':2, 'V3':5,
    'V2':'V1':2, 'V3':2, 'V4':4, 'V5':6,
    'V3':'V1':5, 'V2':2, 'V4':1, 'V6':3,
    'V4':'V2':4, 'V3':1, 'V5':4, 'V6':1,'V7':4,
    'V5':'V2':6, 'V4':4, 'V7':1,
    'V6':'V3':3, 'V4':1, 'V7':2,
    'V7':'V4':4, 'V5':1, 'V6':2


#   dijkstra 算法
def dijkstra_search(graph, start, goal):
    #pass
    frontier = PriorityQueue()
    frontier.put(start, 0)
    came_from =
    came_from[start] = None
    cost_so_far =
    cost_so_far[start] = 0
    
    while not frontier.empty():
        current = frontier.get()
        
        if current == goal:
            break
        for next in graph.neighbors(current):
            new_cost = cost_so_far[current] + graph.cost(current,next)
            if next not in cost_so_far or new_cost < cost_so_far[next]:
                cost_so_far[next]=new_cost
                priority = new_cost
                frontier.put(next, priority)
                came_from[next] = current
    return came_from, cost_so_far


def reconstruct_path(came_from, start, goal):
    current = goal
    path = [current]
    while current != start:
        current = came_from[current]
        path.append(current)
    path.reverse()
    return path

c, f = dijkstra_search(target_graph, 'V1', 'V7')
path=reconstruct_path(c,'V1','V7')
print("最短路径为：".format(path))
#==============================================================================
# 最短路径为：['V1', 'V2', 'V3', 'V4', 'V6', 'V7']
#==============================================================================

最短路径算法之Dijkstra算法

参考：《大话数据结构》

 

这是一个按照路径长度递增的次序产生最短路径的算法。它并不是一次求出源点到目标点的最短路径，而是一步步求出它们之间顶点的最短路径，过程中都是基于已经求出的最短路径的基础上，求得更远顶点的最短路径，最终得到想要的结果。

#define MAXVEX 9
#define INFINITY 
typedef int PathMatrix[MAXVEX]
typedef int ShortPathTable[MAXVEX]

void ShortestPath_Dijkstra(MGraph G, int v0, PathMatrix *p, ShortPathTable *D)
{
    int v,w,k,min;
    int final[MAXVEX]; //final[w]=1表示求得顶点v0到vw的最短路径
    for(v=0;v<G.numVertexes;v++)
    {
        final[v] = 0; //全部顶点初始化为未知最短路径状态
        (*D)[v] = G.matrix[v0][v]; //将与v0点有连线的顶点加上权值
        (*p)[v] = 0; //初始化路径数组p为0
    }
    (*D)[v0] = 0; //v0至v0路径为0
    final[v0] = 1; //v0至v0不需要求路径
    /*开始主循环，每次求得v0到某个v顶点的最短路径*/
    for(v=1;v<G.numVertexes;v++)
    {
        min = INFINITY;
        for(w=0;w<G.numVertexes;w++) //寻找离v0最近的顶点
        {
            if(!final[w] && (*D)[w] < min)
            {
                k = w;
                min = (*D)[w];
            }
        }
    }
    final[k] = 1; //将目前找到的最近的顶点置为1
    for(w=0;w<G.numVertexes;w++)
    {
        if(!final[w] && (min + G.matrix[k][w])<(*D)[w]) //如果警告v顶点的路径比现在这条路径的长度短的话
        {
            (*D)[w] = min + G.matrix[k][w];
            (*p)[w] = k;
        }
    }    
}