Algorithms Part 1-Question 5- Dijkstra’s shortest-path-最短路徑演算法

Algorithms:
Design and Analysis, Part 1   最短路徑演算法題目要求

  本次要求對於一個200個點的無向圖應用著名的Dijkstra演算法求最短路徑。可以選擇用heap來計算，速度更快。我採用python語言，用了多個字典來加快速度。

In this programming problem you'll code up Dijkstra's shortest-path algorithm.
Download the text file here. (Right click and save link as).
The file contains an adjacency list representation of an undirected weighted graph with 200 vertices labeled 1 to 200. Each row consists of the node tuples that are adjacent to that particular vertex along with the length of that edge. For example, the 6th
row has 6 as the first entry indicating that this row corresponds to the vertex labeled 6. The next entry of this row "141,8200" indicates that there is an edge between vertex 6 and vertex 141 that has length 8200. The rest of the pairs of this row indicate
the other vertices adjacent to vertex 6 and the lengths of the corresponding edges.

Your task is to run Dijkstra's shortest-path algorithm on this graph, using 1 (the first vertex) as the source vertex, and to compute the shortest-path distances between 1 and every other vertex of the graph. If there is no path between a vertex v and vertex
1, we'll define the shortest-path distance between 1 and v to be 1000000.

You should report the shortest-path distances to the following ten vertices, in order: 7,37,59,82,99,115,133,165,188,197. You should encode the distances as a comma-separated string of integers. So if you find that all ten of these vertices except 115 are at
distance 1000 away from vertex 1 and 115 is 2000 distance away, then your answer should be 1000,1000,1000,1000,1000,2000,1000,1000,1000,1000. Remember the order of reporting DOES MATTER, and the string should be in the same order in which the above ten vertices
are given. Please type your answer in the space provided.

IMPLEMENTATION NOTES: This graph is small enough that the straightforward O(mn) time implementation of Dijkstra's algorithm should work fine. OPTIONAL: For those of you seeking an additional challenge, try implementing the heap-based version. Note this requires
a heap that supports deletions, and you'll probably need to maintain some kind of mapping between vertices and their positions in the heap.

  演算法實現

  這個演算法比較簡單，跟著ppt走，實現要求的資料結構，把虛擬碼翻譯一下就ok了。當然最後要選擇題目要求的點的距離，使用list comprehension可以加快這一過程。

  整個演算法實現中，全部以字典實現，速度較快。

  代碼如下：

f=open('dijkstraData.txt','r')vertices=dict()for line in f.readlines():    tmp=line.split()    try:        vertices[int(tmp[0])]={int(x.split(',')[0]):int(x.split(',')[1]) for x in tmp[1:]}    except:        print('error'+tmp)f.closelength=200exped=[1]noexped=list(range(2,length+1))distance={x:0 for x in range(1,length+1)}while len(exped)<length:    maxlimit=1000000    tmpdist=maxlimit    lenvw=tmpdist    for v in exped:        for w in noexped:            if w in vertices[v].keys():                lenvw=distance[v]+vertices[v][w]                if lenvw<tmpdist:                    tmpdist=lenvw                    tmpv=v                    tmpw=w    if tmpdist==maxlimit:break    exped.append(tmpw)    noexped.remove(tmpw)    distance[tmpw]=tmpdistprint('distance is')print(distance)print('max distance is: '+str(max(distance.values())))for ind in [7,37,59,82,99,115,133,165,188,197]:    print(str(ind)+' is : '+str(distance[ind]))

  在對大型資料進行處理前，不妨採用小的資料集進行測試。

  這是測試使用的txt檔案：

1 2,7 3,9 6,142 1,7 3,10 4,153 1,9 2,10 4,11 6,24 2,15 3,11 5,65 4,6 6,96 1,14 3,2 5,9

  計算結果是：

distance is{1: 0, 2: 7, 3: 9, 4: 20, 5: 20, 6: 11}max distance is: 20