[1575. Count All Possible Routes](https://leetcode.com/problems/count-all-possible-routes/)

### 題目描述

You are given an array of **distinct** positive integers locations where `locations[i]` represents the position of city `i`. You are also given integers `start`, `finish` and `fuel` representing the starting city, ending city, and the initial amount of fuel you have, respectively.

At each step, if you are at city `i`, you can pick any city `j` such that `j != i` and `0 <= j < locations.length` and move to city `j`. Moving from city `i` to city `j` reduces the amount of fuel you have by `|locations[i] - locations[j]|`. Please notice that `|x|` denotes the absolute value of `x`.

Notice that `fuel` **cannot** become negative at any point in time, and that you are **allowed** to visit any city more than once (including `start` and `finish`).

Return *the count of all possible routes from* `start` *to* `finish`. Since the answer may be too large, return it modulo 10^9^ + 7.

### 範例

**Example 1:**

Input: locations = [2,3,6,8,4], start = 1, finish = 3, fuel = 5
Output: 4
Explanation: The following are all possible routes, each uses 5 units of fuel:
1 -> 3
1 -> 2 -> 3
1 -> 4 -> 3
1 -> 4 -> 2 -> 3

**Example 2:**

Input: locations = [4,3,1], start = 1, finish = 0, fuel = 6
Output: 5
Explanation: The following are all possible routes:
1 -> 0, used fuel = 1
1 -> 2 -> 0, used fuel = 5
1 -> 2 -> 1 -> 0, used fuel = 5
1 -> 0 -> 1 -> 0, used fuel = 3
1 -> 0 -> 1 -> 0 -> 1 -> 0, used fuel = 5

**Example 3:**

Input: locations = [5,2,1], start = 0, finish = 2, fuel = 3
Output: 0
Explanation: It is impossible to get from 0 to 2 using only 3 units of fuel since the shortest route needs 4 units of fuel.

**Constraints**:

* 2 <= `locations.length` <= 100
* 1 <= `locations[i]` <= 10^9^
* All integers in `locations` are **distinct**.
* 0 <= `start`, `finish` < `locations.length`
* 1 <= `fuel` <= 200

### 解答
#### Python

class Solution:
    def countRoutes(self, locations: List[int], start: int, finish: int, fuel: int) -> int:
        MOD = 1000000007
        n = len(locations)

        @cache
        def dfs(cur, remain):
            total = 1 if cur == finish else 0
            for child in range(n):
                distance = abs(locations[child] - locations[cur])
                if child == cur or distance > remain:
                    continue
                total += dfs(child, remain - distance)
            return total % MOD
        
        return dfs(start, fuel)

> [name=Yen-Chi Chen][time=Sun, Jun 25, 2023]

### Reference

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