題目大意：
輸入了醫療隊的編號和他們之間相連接需要的條數
輸出指定的兩個醫療隊的最短路徑的條數，並且在最短路徑中能聚集的醫療隊的數量
例子：0到2的最短路徑可以是0-2.0-1-2.他們都是2的距離，但是0-1-2能聚集1+2+1的醫療隊，輸出4
用到深度優先算法：儘可能深的搜索，當節點所在的邊被探尋過或者不滿足條件時，回朔到上一步，反覆進行該過程直到所有的點遍歷
普適模板：
void dfs(int step):
{
if 判斷邊界條件
{
}
for 嘗試每一種可能
{
if判斷滿足條件
標記
繼續下一步的dfs
恢復初始狀態
}
}
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)] #橫縱座標爲兩個點，數值爲距離
visit = [0 for i in range(cities_num)] #訪問節點列表
min_roads = inf #最短路徑
min_roads_count = 0 #最短路徑條數
max_resue_sum =0 #最大救援隊個數
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): #如果小於當前的road數則更新
min_roads_count = 1
min_roads = road
max_resue_sum = resue
elif(road == min_roads):#如果等於則更新營救隊的最大個數
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)