利用SPFA算法解决PAT A1003
在刷题的时候发现Dijkstra经常会出现超时的情况，一般是使用Dijkstra堆优化或SPFA算法，还有的会使用链式前向星的方法存图都可以减少时间，下面给出SPFA的BFS算法，SPFA算法由Bellman算法演变而来，原理大致就是每当改变一个点的最短路径后，检测与这个点相邻的其他点的最短路径是否改变。
A1003 Emergency (25分)
As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.
Input Specification:
Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (≤500) - the number of cities (and the cities are numbered from 0 to N−1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.
Output Specification:
For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather. All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.
Sample Input:
5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1
Sample Output:
2 4大致题意：
第一行分别给出点数，边数，起点编号和终点编号
第二行给出每个点的权重
之后给出每个边的起点终点以及边的长度
现在求起点到终点的最短距离，如果最短距离有多条输出点权之和最大的
注意点
SPFA和Bellman算法都有个共同点就是在求最短路径条数时，如果发现还有其他路径同样距离最短时要重新按照pre[]动态数组进行求和
题解
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