JAVA高级面试进阶训练营视频教程

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　　遇到一个算法题目，描述如下：

　　

　　对图结构有了解的不难发现，这是经典的求图的最短路径问题。以下是python代码：

def findMin(row):
    minL = max(row)
    for i in row:
        if i != -1 and minL > i:
            minL = i
    return minL
def initRow(row, plus):
    r = []
    for i in row:
        if i != -1:
            i += plus
        r.append(i)
    return r

def getMinLen(table, e, t):
    count = len(table) - 1
    startPoint = 1
    #记录原点到各点最短距离 初始值为-1，即不可达
    lenRecord = list((-1 for i in range(count+1)))
    lenRecord[startPoint] = 0
    #记录每次循环的起点
    points = [startPoint]
    #已得到最短距离的点
    visited = set()
    while len(points)>0:
        #当前起点
        curPoint = points.pop()
        #原点到当前起点的距离
        curLen = lenRecord[curPoint]
        #当前起点到各点的距离
        curList = initRow(table[curPoint], t)
        #当前起点到各点的最短距离
        curMin = findMin(curList)
        visited.add(curPoint)
        idx = 0
        while idx<count:
            idx += 1
            #当前点不可达或到当前点的最短距离已计算出 则跳过
            if curList[idx] == -1 or idx in visited:
                continue
            #记录距离当前起点最近的点作为下次外层循环的起点
            if curList[idx] == curMin:
                points.append(idx)
            #如果从原点经当前起点curPoint到目标点idx的距离更短，则更新
            if lenRecord[idx] == -1 or lenRecord[idx] > (curLen+curList[idx]):
                lenRecord[idx] = curLen+curList[idx]
    return lenRecord[e]

def processInput():
    pointCnt, roadCnt, jobCnt = (int(x) for x in raw_input().split())
    table = []
    for i in range(pointCnt+1):
        table.append([-1] * (pointCnt+1))
    for i in range(roadCnt):
        (x, y, w) = (int(n) for n in raw_input().split())
        if table[x][y] == -1 or table[x][y] > w:
            table[x][y] = w
            table[y][x] = w
    res = []
    for i in range(jobCnt):
        e, t = (int(x) for x in raw_input().split())
        res.append(getMinLen(table, e, t))
    for i in res:
        print(i)

processInput()