题目
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
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
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
解答
#求图上两点间的最短路径，并求最短路径上的最大权值和，如果仅一条最短路径，即为该条路径的权值和
INT_MAX=100000
mind =INT_MAX
maxw=0 #最短路径权值和的最大值
cnt=1 #最短路径条数
N,M,c1,c2=[int(x) for x in input().split(' ')]
wei=[int(x) for x in input().split(' ')]
visit=[-1]*N
#road 为列表的列表，不能用road=[tem[:]]*N，这样各行实际指向同一列表。要避免这种浅复制
#road[i][j]=INT_MAX，代表图上不存在i和j间的边
tem=[INT_MAX]*N
road=[]
for i in range(N):
road.append(tem[:])
#读入各边的长度
for r in range(M):
i,j,dist=[int(x) for x in input().split(' ')]
road[i][j]=dist
road[j][i] = dist
def dfs( p, end, dist, weit):
global mind,cnt,maxw,N,wei,visit
if(p==end):
if(dist<mind):
cnt = 1
mind=dist
maxw = weit
elif(dist==mind):
cnt+=1
if(maxw<weit):
maxw = weit
return 0
if (dist > mind): #如果该路径的长度已经超过已知的最短路径了的话，就没必要继续迭代了
return 1
for i in range(N):
if(visit[i] == -1 and road[p][i] != INT_MAX):
visit[i] = 1
dfs(i, end, dist + road[p][i], weit + wei[i])
visit[i] = -1
visit[c1] = 1
dfs(c1, c2, 0, wei[c1])
print (cnt, maxw)
