diff --git a/0778.Swim-in-Rising-Water/memo.md b/0778.Swim-in-Rising-Water/memo.md
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+# 778. Swim in Rising Water
+
+## step1
+自力で二分探索を思いつくことができた。計算量 O(n^2 logn)。
+
+時間 t と grid の値の不等号の向きを間違えて時間を溶かした。21m。
+
+## step2
+
+> Use either Dijkstra's, or binary search for the best time T for which you can reach the end if you only step on squares at most T.
+
+Dijkstraを使うのか。和ではなくmaxをcostの計算に使えばよい。こちらも計算量O(n^2 logn)。
+
+BFSを毎回行う二分探索よりも定数倍速い。
+
+---
+https://leetcode.com/problems/swim-in-rising-water/solutions/7252184/swim-in-rising-water-3-approach-editoria-8zp6/?envType=problem-list-v2&envId=7p55wqm
+
+Union-Find + Kruskal
+
+## step3
+TODO: Kruskal法を書く
diff --git a/0778.Swim-in-Rising-Water/step1_binary_search.py b/0778.Swim-in-Rising-Water/step1_binary_search.py
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+++ b/0778.Swim-in-Rising-Water/step1_binary_search.py
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+class Solution:
+    def swimInWater(self, grid: list[list[int]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        if len(grid) != len(grid[0]):
+            raise ValueError("invalid grid")
+
+        n = len(grid)
+
+        def can_reach_within(t):
+            if grid[0][0] > t:
+                return False
+
+            reachable = [(0, 0)]
+            seen = set((0, 0))
+            while reachable:
+                next_reachable = []
+                for r, c in reachable:
+                    if r == n - 1 and c == n - 1:
+                        return True
+                    for r_next, c_next in ((r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)):
+                        if not (0 <= r_next < n and 0 <= c_next < n):
+                            continue
+                        if (r_next, c_next) in seen or grid[r_next][c_next] > t:
+                            continue
+                        seen.add((r_next, c_next))
+                        next_reachable.append((r_next, c_next))
+                reachable = next_reachable
+
+            return False
+
+
+        left = 0
+        right = n * n - 1
+        while left < right:
+            mid = (left + right) // 2
+            if not can_reach_within(mid):
+                left = mid + 1
+            else:
+                right = mid
+
+        return left
+
+
diff --git a/0778.Swim-in-Rising-Water/step2_dijkstra.py b/0778.Swim-in-Rising-Water/step2_dijkstra.py
new file mode 100644
index 0000000..94de637
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+++ b/0778.Swim-in-Rising-Water/step2_dijkstra.py
@@ -0,0 +1,27 @@
+import heapq
+
+class Solution:
+    def swimInWater(self, grid: list[list[int]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        if len(grid) != len(grid[0]):
+            raise ValueError("invalid grid")
+
+        n = len(grid)
+        heap = [(grid[0][0], 0, 0)]
+        costs = [n * n] * (n * n)
+        costs[0] = grid[0][0]
+
+        while heap:
+            cost, r, c = heapq.heappop(heap)
+            if r == n - 1 and c == n - 1:
+                return cost
+
+            for r_next, c_next in ((r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)):
+                if not (0 <= r_next < n and 0 <= c_next < n):
+                    continue
+                cost_next = max(cost, grid[r_next][c_next])
+                if cost_next < costs[r_next * n + c_next]:
+                    costs[r_next * n + c_next] = cost_next
+                    heapq.heappush(heap, (cost_next, r_next, c_next))
+
diff --git a/0778.Swim-in-Rising-Water/step2_dijkstra_revised.py b/0778.Swim-in-Rising-Water/step2_dijkstra_revised.py
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index 0000000..39354cb
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+++ b/0778.Swim-in-Rising-Water/step2_dijkstra_revised.py
@@ -0,0 +1,27 @@
+import heapq
+
+class Solution:
+    def swimInWater(self, grid: list[list[int]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        if len(grid) != len(grid[0]):
+            raise ValueError("invalid grid")
+
+        n = len(grid)
+        candidate_cells = [(grid[0][0], 0, 0)]
+        costs = [n * n] * (n * n)
+        costs[0] = grid[0][0]
+
+        while candidate_cells:
+            cost, r, c = heapq.heappop(candidate_cells)
+            if r == n - 1 and c == n - 1:
+                return cost
+
+            for r_next, c_next in ((r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)):
+                if not (0 <= r_next < n and 0 <= c_next < n):
+                    continue
+                cost_next = max(cost, grid[r_next][c_next])
+                if cost_next < costs[r_next * n + c_next]:
+                    costs[r_next * n + c_next] = cost_next
+                    heapq.heappush(candidate_cells, (cost_next, r_next, c_next))
+
diff --git a/0778.Swim-in-Rising-Water/step2_union_find.py b/0778.Swim-in-Rising-Water/step2_union_find.py
new file mode 100644
index 0000000..124e862
--- /dev/null
+++ b/0778.Swim-in-Rising-Water/step2_union_find.py
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+class UnionFind:
+    def __init__(self, n):
+        self.parent = list(range(n))
+
+    def find(self, i):
+        if self.parent[i] == i:
+            return i
+        self.parent[i] = self.find(self.parent[i])
+        return self.parent[i]
+
+    def union(self, i, j):
+        root_i = self.find(i)
+        root_j = self.find(j)
+        if root_i != root_j:
+            self.parent[root_i] = root_j
+            return True
+        return False
+
+
+class Solution:
+    def swimInWater(self, grid: list[list[int]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        if len(grid) != len(grid[0]):
+            raise ValueError("invalid grid")
+        if len(grid) == 1:
+            return grid[0][0]
+
+        n = len(grid)
+
+        edges = []
+        for r in range(n):
+            for c in range(n):
+                if r + 1 < n:
+                    cost = max(grid[r][c], grid[r + 1][c])
+                    edges.append((cost, r * n + c, (r + 1)* n + c))
+                if c + 1 < n:
+                    cost = max(grid[r][c], grid[r][c + 1])
+                    edges.append((cost, r * n + c, r * n + c + 1))
+
+        edges.sort()
+
+        uf = UnionFind(n * n)
+
+        for cost, u, v in edges:
+            uf.union(u, v)
+            if uf.find(0) == uf.find((n - 1) * n + n - 1):
+                return cost
+
+        return -1