Question

我正在努力解决我在做作业时遇到的问题，并且真的觉得我正在过度思考这个算法，希望有人在这里推动我朝着正确的方向前进。

我将获得一个输入txt文件，如下所示：

1    // n number of graphs
4    // n number of vertices for graph 1
4    // n number of edges for graph 1
1 2  // edges given in pairs
2 3
2 4
3 4

我应该使用这些数据来创建代表图形的n个邻接矩阵。然后我需要对邻接矩阵中的数据实现3种方法：

findLongestPath（）将返回图表中最长的路径
findShortestPath（）将返回图表中的最短路径
totalNumberPaths（）将返回图表中不同数量的路径

我很难实现前两部分。这就是我到目前为止所做的：

def main():


numGraphs = input()

for x in xrange(0, numGraphs):
    numVerts = input()
    numEdges = input()
    adjMat = [[0 for x in xrange(numVerts)] for x in xrange(numVerts)] 
    for x in xrange(0, numEdges):
        edges = raw_input()
        i, padding, j = edges.rpartition(" ")

        i = int(i)
        j = int(j)

        i -= 1
        j -= 1

        adjMat[i][j] = 1


    numPaths = [0 for x in xrange(numVerts)]
    numPaths[0] = 1 

    longest_path = 1
    shortest_path = numVerts

    for i in xrange(0, numVerts):
        current_path = 0
        for j in xrange(0, numVerts):
            if adjMat[i][j] == 1:
                numPaths[j] += numPaths[i]
                current_path += 1

        if current_path > longest_path:
            longest_path = current_path
        if current_path < shortest_path:
            shortest_path = current_path

    print "shortest: %d, longest: %d, total %d" % (shortest_path, longest_path, numPaths[numVerts-1])

 if __name__ == "__main__":
     main()

显然，当它遇到一行0时，shortest_path会更新为0并且不起作用。另外，当初始化为0时，它将无法工作。如果我可以获得一些伪代码或者可能有助于使用更长或更短的方法，我确信我可以写相反的内容，或者我可能完全偏离基础。

感谢您的任何意见。

编辑：

所以我想通了。这是我完成的代码，以防任何人遇到类似问题并需要帮助。

numGraphs = input()

for x in xrange(0, numGraphs):
    numVerts = input()
    numEdges = input()
    adjMat = [[0 for x in xrange(numVerts)] for x in xrange(numVerts)] 
    for x in xrange(0, numEdges):
        edges = raw_input()
        i, padding, j = edges.rpartition(" ")

        i = int(i)
        j = int(j)

        i -= 1
        j -= 1

        adjMat[i][j] = 1


    numPaths = [0 for x in xrange(numVerts)]
    numPaths[0] = 1 

    currentPath = [0 for x in xrange(numVerts)]
    maxPath = 1
    minPath = numVerts -1

    for i in xrange(0, numVerts):
        for j in xrange(1, numVerts):
            if adjMat[i][j] == 1:
                numPaths[j] += numPaths[i]
                currentPath[j-i] += 1
            if (currentPath[j-i] is not 0):
                minPath = currentPath[j-i]
            maxPath = max(currentPath)

    print "shortest: %d, longest: %d, total %d" % (minPath, maxPath, numPaths[numVerts-1])

Answer 1

想出来。这是我的最终解决方案。

numGraphs = input()

for x in xrange(0, numGraphs):
    numVerts = input()
    numEdges = input()
    adjMat = [[0 for x in xrange(numVerts)] for x in xrange(numVerts)] 
    for x in xrange(0, numEdges):
        edges = raw_input()
        i, padding, j = edges.rpartition(" ")

        i = int(i)
        j = int(j)

        i -= 1
        j -= 1

        adjMat[i][j] = 1


    numPaths = [0 for x in xrange(numVerts)]
    numPaths[0] = 1 

    currentPath = [0 for x in xrange(numVerts)]
    maxPath = 1
    minPath = numVerts -1

    for i in xrange(0, numVerts):
        for j in xrange(1, numVerts):
            if adjMat[i][j] == 1:
                numPaths[j] += numPaths[i]
                currentPath[j-i] += 1
            if (currentPath[j-i] is not 0):
                minPath = currentPath[j-i]
            maxPath = max(currentPath)

    print "shortest: %d, longest: %d, total %d" % (minPath, maxPath, numPaths[numVerts-1])

使用邻接矩阵（Python或伪代码）在拓扑排序的非加权有向无环图中的最短和最长路径

1 个答案: