|
| 1 | +# 并查集 |
| 2 | + |
| 3 | +用于处理不相交集合 (disjoint sets) 合并及查找的问题。 |
| 4 | + |
| 5 | +### [redundant-connection](https://leetcode-cn.com/problems/redundant-connection/) |
| 6 | + |
| 7 | +```Python |
| 8 | +class Solution: |
| 9 | + def findRedundantConnection(self, edges: List[List[int]]) -> List[int]: |
| 10 | + |
| 11 | + parent = list(range(len(edges) + 1)) |
| 12 | + |
| 13 | + def find(x): |
| 14 | + if parent[parent[x]] != parent[x]: |
| 15 | + parent[x] = find(parent[x]) |
| 16 | + return parent[x] |
| 17 | + |
| 18 | + def union(x, y): |
| 19 | + px, py = find(x), find(y) |
| 20 | + if px == py: |
| 21 | + return False |
| 22 | + parent[px] = py |
| 23 | + return True |
| 24 | + |
| 25 | + for u, v in edges: |
| 26 | + if not union(u, v): |
| 27 | + return [u, v] |
| 28 | +``` |
| 29 | + |
| 30 | +### [accounts-merge](https://leetcode-cn.com/problems/accounts-merge/) |
| 31 | + |
| 32 | +```Python |
| 33 | +class Solution: |
| 34 | + def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]: |
| 35 | + |
| 36 | + parent = [] |
| 37 | + |
| 38 | + def find(x): |
| 39 | + if parent[parent[x]] != parent[x]: |
| 40 | + parent[x] = find(parent[x]) |
| 41 | + return parent[x] |
| 42 | + |
| 43 | + def union(x, y): |
| 44 | + parent[find(x)] = find(y) |
| 45 | + return |
| 46 | + |
| 47 | + email2name = {} |
| 48 | + email2idx = {} |
| 49 | + i = 0 |
| 50 | + for acc in accounts: |
| 51 | + for email in acc[1:]: |
| 52 | + email2name[email] = acc[0] |
| 53 | + if email not in email2idx: |
| 54 | + parent.append(i) |
| 55 | + email2idx[email] = i |
| 56 | + i += 1 |
| 57 | + union(email2idx[acc[1]], email2idx[email]) |
| 58 | + |
| 59 | + result = collections.defaultdict(list) |
| 60 | + for email in email2name: |
| 61 | + result[find(email2idx[email])].append(email) |
| 62 | + |
| 63 | + return [[email2name[s[0]]] + sorted(s) for s in result.values()] |
| 64 | +``` |
| 65 | + |
| 66 | + |
| 67 | + |
| 68 | +### Kruskal's algorithm |
| 69 | + |
| 70 | +### [minimum-risk-path](https://www.lintcode.com/problem/minimum-risk-path/description) |
| 71 | + |
| 72 | +> 地图上有 m 条无向边,每条边 (x, y, w) 表示位置 m 到位置 y 的权值为 w。从位置 0 到 位置 n 可能有多条路径。我们定义一条路径的危险值为这条路径中所有的边的最大权值。请问从位置 0 到 位置 n 所有路径中最小的危险值为多少? |
| 73 | + |
| 74 | +**图森面试真题**。最小危险值为最小生成树中 0 到 n 路径上的最大边权。 |
| 75 | + |
| 76 | +```Python |
| 77 | +# Kruskal's algorithm |
| 78 | +class Solution: |
| 79 | + def getMinRiskValue(self, N, M, X, Y, W): |
| 80 | + |
| 81 | + # Kruskal's algorithm with union-find to construct MST |
| 82 | + parent = list(range(N + 1)) |
| 83 | + |
| 84 | + def find(x): |
| 85 | + if parent[parent[x]] != parent[x]: |
| 86 | + parent[x] = find(parent[x]) |
| 87 | + return parent[x] |
| 88 | + |
| 89 | + def union(x, y): |
| 90 | + px, py = find(x), find(y) |
| 91 | + if px != py: |
| 92 | + parent[px] = py |
| 93 | + return True |
| 94 | + else: |
| 95 | + return False |
| 96 | + |
| 97 | + edges = sorted(zip(W, X, Y)) |
| 98 | + |
| 99 | + MST_edges = [] |
| 100 | + for edge in edges: |
| 101 | + if union(edge[1], edge[2]): |
| 102 | + MST_edges.append(edge) |
| 103 | + if find(0) == find(N): |
| 104 | + break |
| 105 | + |
| 106 | + MST = collections.defaultdict(list) |
| 107 | + target = find(0) |
| 108 | + for w, u, v in MST_edges: |
| 109 | + if find(u) == target and find(v) == target: |
| 110 | + MST[u].append((v, w)) |
| 111 | + MST[v].append((u, w)) |
| 112 | + |
| 113 | + # dfs to search route from 0 to n |
| 114 | + dfs = [(0, None, float('-inf'))] |
| 115 | + while dfs: |
| 116 | + v, p, max_w = dfs.pop() |
| 117 | + for n, w in MST[v]: |
| 118 | + cur_max_w = max(max_w, w) |
| 119 | + if n == N: |
| 120 | + return cur_max_w |
| 121 | + if n != p: |
| 122 | + dfs.append((n, v, cur_max_w)) |
| 123 | +``` |
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