题目大意：
输入了医疗队的编号和他们之间相连接需要的条数
输出指定的两个医疗队的最短路径的条数，并且在最短路径中能聚集的医疗队的数量
例子：0到2的最短路径可以是0-2.0-1-2.他们都是2的距离，但是0-1-2能聚集1+2+1的医疗队，输出4
用到深度优先算法：尽可能深的搜索，当节点所在的边被探寻过或者不满足条件时，回朔到上一步，反复进行该过程直到所有的点遍历
普适模板：
void dfs(int step):
{
if 判断边界条件
{
}
for 尝试每一种可能
{
if判断满足条件
标记
继续下一步的dfs
恢复初始状态
}
}
inf = float('inf')
cities_num, roads_num, city_start, city_end = map(int,input().split())
rescue_teams = list(map(int,input().split()))
roads = [[inf for i in range(cities_num)] for i in range(cities_num)] #横纵座标为两个点，数值为距离
visit = [0 for i in range(cities_num)] #访问节点列表
min_roads = inf #最短路径
min_roads_count = 0 #最短路径条数
max_resue_sum =0 #最大救援队个数
for i in range(roads_num):
city1, city2, road = map(int,input().split())
roads[city1][city2] = road
roads[city2][city1] = road
def dfs(start, end, road, resue):
global min_roads, min_roads_count, max_resue_sum,roads_num,rescue_teams,visit
if(start == end):
if(road < min_roads): #如果小于当前的road数则更新
min_roads_count = 1
min_roads = road
max_resue_sum = resue
elif(road == min_roads):#如果等於则更新营救队的最大个数
min_roads_count += 1
if(max_resue_sum < resue):
max_resue_sum = resue
return 0
if(road > min_roads):
return 1
for i in range(cities_num):
if(visit[i] == 0 and roads[start][i] != inf):
visit[i] = 1
dfs(i, end, road+roads[start][i], resue+rescue_teams[i])
visit[i] = 0
visit[city_start] = 1
dfs(city_start,city_end,0,rescue_teams[city_start])
print(min_roads_count,max_resue_sum)