feat: todo

backlog/dp/bus-fare.py

@@ -0,0 +1,61 @@
+class Solution:
+    def solve(self, days):
+        # n = len(days)
+        # prices = [2, 7, 25]
+        # durations = [1, 7, 30]
+        # dp = [float("inf")] * (n + 1)
+        # # dp[i] 表示截止第 i + 1 天（包括）需要多少钱，因此答案就是 dp[n]
+        # dp[0] = 0
+
+        # for i in range(1, n + 1):
+        #     for j in range(i, n + 1):
+        #         for price, duration in zip(prices, durations):
+        #             # [i-1, j-1] 闭区间  -> dp [i,j]  ->  dp[i-1]
+        #             if days[j - 1] - days[i - 1] + 1 <= duration:
+        #                 dp[j] = min(dp[j], dp[i - 1] + price)
+        # return dp[-1]
+
+        # m*n^2 => m*nlogn -> m*n   m = 3 n = len(days)
+
+        # n = len(days)
+        # prices = [2, 7, 25]
+        # durations = [1, 7, 30]
+
+        # @lru_cache(None)
+        # def dp(i):
+        #     if i >= n:
+        #         return 0
+        #     return min([price + dp(bisect.bisect_left(days, days[i] + duration)) for price, duration in zip(prices, durations)])
+
+        # return dp(0)
+
+        # n = len(days)
+        # prices = [2, 7, 25]
+        # durations = [1, 7, 30]
+        # dp = [float("inf")] * (n + 1)
+        # # dp[i] 表示截止第 i + 1 天（包括）需要多少钱，因此答案就是 dp[n]
+        # dp[0] = 0
+
+        # for i in range(1, n + 1):
+        #     for j in range(i, n + 1):
+        #         for price, duration in zip(prices, durations):
+        #             if days[j - 1] - days[i - 1] + 1 <= duration:
+        #                 dp[j] = min(dp[j], dp[i - 1] + price)
+        #             elif price == 25:
+        #                 break
+
+        # return dp[-1]
+        prices = [2, 7, 25]
+        durations = [1, 7, 30]
+        n = len(days)
+        m = len(prices)
+        dp = [float("inf")] * (n + 1)
+        dp[0] = 0
+        pointers = [0] * m
+        # 上面 dp 的问题在于 prices 指针不断回溯，实际上没有必要。因为xxxx(上面的 break)，比如上一次 price 为 2 的时候内层（第二层）走到 5（j == 5）了，那么下一次 price 为 2 的时候从 5 开始就行了，前面不用看的，都不满足了。因此可使用一个数组记录指针，并保证指针只前进不回退，这样时间复杂度可减低到 m*n
+        for i in range(1, n + 1):
+            for j in range(m):
+                while days[i - 1] - days[pointers[j]] >= durations[j]:
+                    pointers[j] += 1
+                dp[i] = min(dp[i], dp[pointers[j]] + prices[j])
+        return dp[-1]