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1018. Public Bike Management (30) [Dijkstra算法   DFS]There is a public bike service in Hangzhou City which provides great convenience to the tourists from allover the world. One may rent a bike at any station and return it to any other stations in the city.The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. Astation is said to be in perfect condition if it is exactly half-full. If a station is full or empty, PBMC willcollect or send bikes to adjust the condition of that station to perfect. And more, all the stations on theway will be adjusted as well.When a problem station is reported, PBMC will always choose the shortest path to reach that station. Ifthere are more than one shortest path, the one that requires the least number of bikes sent from PBMCwill be chosen.1. PBMC -> S1 -> S3. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike fromS1 and then take 5 bikes to S3, so that both stations will be in perfect conditions.2. PBMC -> S2 -> S3. This path requires the same time as path 1, but only 3 bikes sent from PBMC andhence is the one that will be chosen.Input Specification:Each input file contains one test case. For each case, the first line contains 4 numbers: Cmax (<= 100),always an even number, is the maximum capacity of each station; N (<= 500), the total number ofstations; Sp, the index of the problem station (the stations are numbered from 1 to N, and PBMC isrepresented by the vertex 0); and M, the number of roads. The second line contains N non-negativenumbers Ci (i=1,…N) where each Ci is the current number of bikes at Si respectively. Then M lines follow,each contains 3 numbers: Si, Sj, and Tij which describe the time Tij taken to move betwen stations Si andSj. All the numbers in a line are separated by a space. www.jingjixx.com/guoji/20210130-1356.htmlOutput Specification:For each test case, print your results in one line. First output the number of bikes that PBMC must send.Then afer one space, output the path in the format: 0->S1->…->Sp. Finally afer another space, outputthe number of bikes that we must take back to PBMC afer the condition of Sp is adjusted to perfect.Note that if such a path is not unique, output the one that requires minimum number of bikes that wemust take back to PBMC. The judge’s data guarantee that such a path is unique.Sample Input:10 3 3 56 7 00 1 10 2 10 3 31 3 12 3 1Sample Output:3 0->2->3 0题⽬⼤意：每个⾃⾏⻋⻋站的最⼤容量为⼀个偶数cmax，如果⼀个⻋站⾥⾯⾃⾏⻋的数量恰好为cmax/ 2，那么称处于完美状态。如果⼀个⻋站容量是满的或者空的，控制中⼼（处于结点0处）就会携带或者从路上收集⼀定数量的⾃⾏⻋前往该⻋站，⼀路上会让所有的⻋站沿途都达到完美。现在给出cmax，⻋站的数量n，问题⻋站sp，m条边，还有距离，求最短路径。如果最短路径有多个，求能带的最少的⾃⾏⻋数⽬的那条。如果还是有很多条不同的路，那么就找⼀个从⻋站带回的⾃⾏⻋数⽬最少的（带回的时候是不调整的）～分析：Dijkstra   DFS。如果只有Dijkstra是不可以的，因为minNeed和minBack在路径上的传递不满⾜最优⼦结构，不是简单的相加的过程，只有在所有路径都确定了之后才能区选择最⼩的need和最⼩的back～Dijkstra求最短路径，dfs求minNeed和minBack和path，dfs的时候模拟⼀遍需要调整的过程，求出最后得到的need和back，与minNeed和minBack⽐较然后根据情况更新path，最后输出minNeed path 和minBack，记得path是从最后⼀个结点⼀直到第⼀个结点的，所以要倒着输出

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