diff --git a/Dynamic_Programming/3538.Merge-Operations-for-Minimum-Travel-Time/3538.Merge-Operations-for-Minimum-Travel-Time.cpp b/Dynamic_Programming/3538.Merge-Operations-for-Minimum-Travel-Time/3538.Merge-Operations-for-Minimum-Travel-Time.cpp
new file mode 100644
index 000000000..45d1a4ca3
--- /dev/null
+++ b/Dynamic_Programming/3538.Merge-Operations-for-Minimum-Travel-Time/3538.Merge-Operations-for-Minimum-Travel-Time.cpp
@@ -0,0 +1,30 @@
+const int INF = INT_MAX / 2;
+int dp[51][11][101];
+
+class Solution {    
+public:
+    int minTravelTime(int l, int n, int K, vector<int>& pos, vector<int>& time) {
+        
+        fill(&dp[0][0][0], &dp[0][0][0]+51*11*101, INT_MAX/2);
+        
+        dp[0][0][time[0]] = 0;
+
+        for (int i=0; i<n; i++)
+            for (int m=0; m<=K; m++)
+                for (int v=0; v<=100; v++)
+                {
+                    int t = 0;                                        
+                    for (int j=i+1; j<n && m+j-i-1<=K; j++)
+                    {
+                        t += time[j];
+                        dp[j][m+j-i-1][t] = min(dp[j][m+j-i-1][t], dp[i][m][v] + v*(pos[j]-pos[i])); 
+                    } 
+                }
+
+        int ans = INT_MAX/2;
+        for (int v = 0; v <= 100; v++) {
+            ans = min(ans, dp[n - 1][K][v]);
+        }
+        return ans;
+    }
+};
diff --git a/Dynamic_Programming/3538.Merge-Operations-for-Minimum-Travel-Time/Readme.md b/Dynamic_Programming/3538.Merge-Operations-for-Minimum-Travel-Time/Readme.md
new file mode 100644
index 000000000..350fb8a8e
--- /dev/null
+++ b/Dynamic_Programming/3538.Merge-Operations-for-Minimum-Travel-Time/Readme.md
@@ -0,0 +1,9 @@
+### 3538.Merge-Operations-for-Minimum-Travel-Time
+
+我们比较容易想到令dp[i][k]表示前进至第i个标记时、且做了k次合并的最短时间。此时为了计算时间，我们需要知道上一处合并点的位置j，以及从j到i的效率（即单位长度所用的时间）。考虑到约束条件`1 <= sum(time) <= 100​​​​​​`，这时一个可以遍历的数字，所以我们就可以将其作为第三个维度。即我们定理dp[i][k][v]表示前进至第i个标记时、恰做了k次合并、且从i开始的效率是v的情况下，所用的最短时间。
+
+写出上述表达式之后，我们发现想要计算dp[i][k][v]，如果只遍历上一个停留点j的位置是不够的，因为从j往后到i的效率也是未知的，无法计算从j到i所花的时间。似乎状态转移陷入了僵局。
+
+其实基于dp[i][k][v]，我们发现对未来的状态进行推断更容易一些。我们同样遍历下一个停留点的位置j，那么从i往后到j所需要的时间是基于v且已知的`t = v*(dist[j]-dist[i])`。此外，从j点往后的效率就是`v2=time[i:j]`的区间和，也是已知的。于是，我们就可以用`dp[i][k][v] + t`来尝试更新`dp[j][k+j-i-1][v2]`的值。
+
+这样的时间复杂度就是`50*18*100*40 = 3e6`数量级，可以通过。
diff --git a/Readme.md b/Readme.md
index 6e30d1b38..1b308391e 100644
--- a/Readme.md
+++ b/Readme.md
@@ -961,6 +961,7 @@
 * ``Infer future from current``   
 [2044.Count-Number-of-Maximum-Bitwise-OR-Subsets](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2044.Count-Number-of-Maximum-Bitwise-OR-Subsets) (M)      
 [2742.Painting-the-Walls](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2742.Painting-the-Walls) (H)      
+[3538.Merge-Operations-for-Minimum-Travel-Time](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/3538.Merge-Operations-for-Minimum-Travel-Time) (H)      
 * ``maximum subarray``   
 [053.Maximum-Subarray](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/053.Maximum-Subarray) (E+)  
 [152.Maximum-Product-Subarray](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/152.Maximum-Product-Subarray) (M+)