There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over 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. A station is said to be in perfect condition if it is exactly half-full. If a station is full or empty, PBMC will collect or send bikes to adjust the condition of that station to perfect. And more, all the stations on the way will be adjusted as well. When a problem station is reported, PBMC will always choose the shortest path to reach that station. If there are more than one shortest path, the one that requires the least number of bikes sent from PBMC will be chosen.
The above figure illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex S is the current number of bikes stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S3 , we have 2 different shortest paths: 1.PBMC -> S1 -> S3 . In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S1 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 and hence 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 of stations;Sp , the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers 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 and Sj . All the numbers in a line are separated by a space. Output Specification: For each test case, print your results in one line. First output the number of bikes that PBMC must send. Then after one space, output the path in the format: 0−>S1 −>⋯−>Sp . Finally after another space, output the number of bikes that we must take back to PBMC after 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 we must take back to PBMC. The judge’s data guarantee that such a path is unique. Sample Input:
10 3 3 5 6 7 0 0 1 1 0 2 1 0 3 3 1 3 1 2 3 1Sample Output:
3 0->2->3 0输入的第一个值Cmax是一个车站的最多自行车存量，当一个车站里面的自行车的数量刚好是满的一般，那么称为完美状态 从控制中心出发去往目的车站（Sp），一路上经过的车站，如果车站的车多于完美状态，就带走一些车，如果车站的车少于完美状态，就补上一些车。但是一路上的车站自给自足的可能性很低，因此需要从控制中心带一些车走，也需要最后带回一些车。 现在题目就是要求得到一个最短路径，如果最短路径有多条，则选择从控制中心带走最少车的那条路径，如果这样还有多条路径，则选择最后带回车最少的那条路径，题目保证没有多条。
最短路径需要Dijkstra，但是只有Dijkstra是不够的。因为带走最少车辆minneed和带回最少车辆minback不满足最优子结构条件，不是简单的相加。也就是说，只有到最后整条路径确定了之后才能选择最小的need和back，在单个结点是无法知道的。 因此，则要求Dijkstra可以得到所有的最短路径，这里选择使用vector数组保存顶点的所有前驱顶点，如题所示的图，其最短路径的数组就为：
pre[0]: pre[1]:0 pre[2]:0 pre[3]:1、2然后使用DFS，计算每条最短路径所需的need和back，最后按照要求选择理想的路径。其主要思想是通过递归得到完整的temppath，计算该路径的need和back，根据情况保存结果到path。这里temppath和path都使用vector，比数组更方便赋值。但是在判断完一个结点后需将其弹出，具体在代码中显现。 最后，path的路径是从后往前保存的，所以输出时，需要从后往前输出
