Full Tank? 是怎么实现的呢

Question

Description

After going through the receipts from your car trip through Europe this summer, you realised that the gas prices varied between the cities you visited. Maybe you could have saved some money if you were a bit more clever about where you filled your fuel?

To help other tourists (and save money yourself next time), you want to write a program for finding the cheapest way to travel between cities, filling your tank on the way. We assume that all cars use one unit of fuel per unit of distance, and start with an empty gas tank.

Input

The first line of input gives 1 ≤ n ≤ 1000 and 0 ≤ m ≤ 10000, the number of cities and roads. Then follows a line with n integers 1 ≤ pi ≤ 100, where pi is the fuel price in the ith city. Then follow m lines with three integers 0 ≤ u, v < n and 1 ≤ d ≤ 100, telling that there is a road between u and v with length d. Then comes a line with the number 1 ≤ q ≤ 100, giving the number of queries, and q lines with three integers 1 ≤ c ≤ 100, s and e, where c is the fuel capacity of the vehicle, s is the starting city, and e is the goal.

Output

For each query, output the price of the cheapest trip from s to e using a car with the given capacity, or "impossible" if there is no way of getting from s to e with the given car.

Sample Input

5 5
10 10 20 12 13
0 1 9
0 2 8
1 2 1
1 3 11
2 3 7
2
10 0 3
20 1 4
Sample Output

170
impossible

Answer 1

以下回答参考 皆我百晓生、券券喵儿 等免费微信小程序相关内容作答，并由本人整理回复。

# read data
n, m = map(int, input().split())
prices = list(map(int, input().split()))
roads = []
for _ in range(m):
    u, v, d = map(int, input().split())
    roads.append((u, v, d))
q = int(input())

# dp[i][j] means the minimum cost to fill j units of gas at i-th city
dp = [[float('inf')] * n for _ in range(q + 1)]
# dp[0][i] means the minimum cost to go from s to i without any restriction
dp[0][s] = 0
for i in range(1, q + 1):
    for j in range(n):
        # dp[i][j] = min(dp[i-1][k] + dist(j,k) * prices[k])
        for k in range(j + 1, n):
            dp[i][j] = min(dp[i][j], dp[i - 1][k] + roads[j][2] * prices[k])

ans = []
for _ in range(q):
    c, s, e = map(int, input().split())
    if dp[q][e] == float('inf'):
        ans.append("impossible")
    else:
        ans.append(dp[q][e])
print("\n".join(map(str, ans)))