Floyd-Warshall算法python代码

华电北风吹
2017年3月12日

未连接的边需要赋一个初始值，可以把矩阵所有元素值相加再加1。附加功能是计算最短路径的条数。

import sys

def Floyd(Graph,ShortestPath,PathCount):
    NodeNum=len(Graph)

    lastShortestDistance=[[0 for i in range(NodeNum)] for j in range(NodeNum)]
    lastPathCount=[[0 for i in range(NodeNum)] for j in range(NodeNum)]

    currentShortestDistance=[[0 for i in range(NodeNum)] for j in range(NodeNum)]
    currentPathCount=[[0 for i in range(NodeNum)] for j in range(NodeNum)]

    for i in range(NodeNum):
        for j in range(NodeNum):
            lastShortestDistance[i][j]=Graph[i][j]
            if (Graph[i][j]>0) and (Graph[i][j]<10):
                lastPathCount[i][j]=1
            else:
                lastPathCount[i][j]=0
            currentShortestDistance[i][j]=lastShortestDistance[i][j]
            currentPathCount[i][j]=lastPathCount[i][j]

    for k in range(NodeNum):
        for i in range(NodeNum):
            if (i==k):
                continue
            for j in range(NodeNum):
                if (j==k):
                    continue
                if(lastShortestDistance[i][j]continue
                if(lastShortestDistance[i][j]==lastShortestDistance[i][k]+lastShortestDistance[k][j]):
                    currentShortestDistance[i][j]=lastShortestDistance[i][j]
                    currentPathCount[i][j]=lastPathCount[i][j]+lastPathCount[i][k]*lastPathCount[k][j]
                if(lastShortestDistance[i][j]>lastShortestDistance[i][k]+lastShortestDistance[k][j]):
                    currentShortestDistance[i][j]=lastShortestDistance[i][k]+lastShortestDistance[k][j]
                    currentPathCount[i][j]=lastPathCount[i][k]*lastPathCount[k][j]

        lastShortestDistance=currentShortestDistance
        lastPathCount=currentPathCount

    for i in range(NodeNum):
        for j in range(NodeNum):
            ShortestPath[i][j]=currentShortestDistance[i][j]
            PathCount[i][j]=currentPathCount[i][j]

    return None

nodeNum=4
graph=[[0,1,1,10],[1,0,10,10],[1,10,0,1],[10,10,1,0]]
shortestPath=[[0 for i in range(nodeNum)] for j in range(nodeNum)]
pathCount=[[0 for i in range(nodeNum)] for j in range(nodeNum)]

Floyd(graph,shortestPath,pathCount)
print(shortestPath)
print(pathCount)