meeting backup

Author: the-yuyut

紀錄/46 2022 0628 heap 9th end.md

@@ -26,4 +26,120 @@
 * 	[done] HARD	295	Find Median from Data Stream	https://leetcode.com/problems/find-median-from-data-stream
 *   MEDIUM 1631 Path With Minimum Effort https://leetcode.com/problems/path-with-minimum-effort/
 
+
+---
+
+```python=
+
+import heapq
+class Solution:
+    def minimumEffortPath(self, heights: List[List[int]]) -> int:
+        # use dfs + heap
+        # time analysis
+        # operation of heappop and heappush: O(logH), where H is the size of heap
+        # visited all heights: O(N), where N is the size of given heights (H*W)
+        # overall: O(N)*O(logH)
+        
+        H, W = len(heights), len(heights[0])
+        hp = [(0,0,0)]      # diff, i, j
+        effort = 0
+        visited = set()
+
+        while hp:
+            diff, i, j = heapq.heappop(hp)
+            effort = max(effort, diff)
+            print(i,j,heights[i][j], diff, hp) # log
+            if i == H-1 and j == W-1:
+                return effort
+
+            visited.add((i,j))
+            for a, b in [(i+1,j), (i-1,j), (i,j+1), (i,j-1)]:
+                if 0<=a<H and 0<=b<W and (a,b) not in visited:
+                    print(a,b)                 # log
+                    newDiff = abs(heights[a][b]-heights[i][j])
+                    heapq.heappush(hp, (newDiff, a, b))
+        
+        return effort
+```
+
+```text
+[[1,2,2],[3,8,2],[5,3,5]]
+log
+0 0 1 0 []
+1 0
+0 1
+0 1 2 1 [(2, 1, 0)]
+1 1
+0 2
+0 2 2 0 [(2, 1, 0), (6, 1, 1)]
+1 2
+1 2 2 0 [(2, 1, 0), (6, 1, 1)]
+2 2
+1 1
+1 0 3 2 [(3, 2, 2), (6, 1, 1), (6, 1, 1)]
+2 0
+1 1
+2 0 5 2 [(3, 2, 2), (5, 1, 1), (6, 1, 1), (6, 1, 1)]
+2 1
+2 1 3 2 [(3, 2, 2), (5, 1, 1), (6, 1, 1), (6, 1, 1)]
+1 1
+2 2
+2 2 5 2 [(3, 2, 2), (5, 1, 1), (6, 1, 1), (6, 1, 1), (5, 1, 1)]
+
+```
+
+
+
+
 ---
+Y.J. Lee晚上8:14
+比第一個
+Y.J. Lee晚上8:26
+用visited控制拜訪過的格子
+每次拜訪新格子
+一定是使用目前最小的effort
+拜訪新格子的時候
+把新格子跟對應的新effort推進heap裡面
+Y.J. Lee晚上8:30
+總而言之就是
+永遠選最好走的路
+跟走的距離無關
+Y.J. Lee晚上8:34
+比較關鍵的地方
+在更新visited 的時機
+我覺得啦
+Y.J. Lee晚上8:35
+以題目給的example 3為例子
+(3,1)會被加進去heap很多次
+但是到終點都不會pop
+Y.J. Lee晚上8:37
+你看(1,1)其實被加進去很多次
+Y.J. Lee晚上8:39
+222 走到底發現滿難走的
+回頭走3那格
+Y.J. Lee晚上8:41
+這邊是無腦走看看 會產生困難的路線
+困難的路線會跑到heap底部
+yes
+Y.J. Lee晚上8:44
+XD
+Y.J. Lee晚上8:47
+要真的試著做做看比較能理解
+Y.J. Lee晚上8:53
+dijkstra 就是heap
+dijkstra 就是heap+dfs
+時間複雜度我po在comment
+Y.J. Lee晚上8:54
+sry hackmd comment
+L37
+嗯 worst case scenario
+Y.J. Lee晚上8:55
+就很衰 走來走去都越走effort越高
+Y.J. Lee晚上8:57
+可能不是到所有的點啦 但是正比於面積
+而且你走來走去搞不好同個點被加很多遍
+Y.J. Lee晚上8:58
+不會再走 可是會被加進去heap裡面
+Y.J. Lee晚上8:59
+visited最多就是存全部的點 所以確定是O(N)
+heap你沒辦法控制存多少東西進去 但是是正比於面積 所以heapsize ~= N