|
| 1 | +# 最小生成树 |
| 2 | + |
| 3 | +### [minimum-risk-path](https://www.lintcode.com/problem/minimum-risk-path/description) |
| 4 | + |
| 5 | +> 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? |
| 6 | + |
| 7 | +**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。 |
| 8 | + |
| 9 | +```Python |
| 10 | +# Kruskal's algorithm |
| 11 | +class Solution: |
| 12 | + def getMinRiskValue(self, n, m, x, y, w): |
| 13 | + |
| 14 | + # Kruskal's algorithm with union-find to construct MST |
| 15 | + parent = list(range(n + 1)) |
| 16 | + |
| 17 | + def find(x): |
| 18 | + if parent[parent[x]] != parent[x]: |
| 19 | + parent[x] = find(parent[x]) |
| 20 | + return parent[x] |
| 21 | + |
| 22 | + def union(x, y): |
| 23 | + px, py = find(x), find(y) |
| 24 | + if px != py: |
| 25 | + parent[px] = py |
| 26 | + return True |
| 27 | + else: |
| 28 | + return False |
| 29 | + |
| 30 | + edges = sorted(zip(x, y, w), key=lambda x: x[2]) |
| 31 | + |
| 32 | + mst_edges = [] |
| 33 | + for edge in edges: |
| 34 | + if union(edge[0], edge[1]): |
| 35 | + mst_edges.append(edge) |
| 36 | + if find(0) == find(n): |
| 37 | + break |
| 38 | + |
| 39 | + mst = collections.defaultdict(list) |
| 40 | + target = find(0) |
| 41 | + for u, v, r in mst_edges: |
| 42 | + if find(u) == target and find(v) == target: |
| 43 | + mst[u].append((v, r)) |
| 44 | + mst[v].append((u, r)) |
| 45 | + |
| 46 | + # dfs to search route from 0 to n |
| 47 | + dfs = [(0, None, float('-inf'))] |
| 48 | + while dfs: |
| 49 | + v, p, max_risk = dfs.pop() |
| 50 | + for a, r in mst[v]: |
| 51 | + cur_max = max(max_risk, r) |
| 52 | + if a == n: |
| 53 | + return cur_max |
| 54 | + if a != p: |
| 55 | + dfs.append((a, v, cur_max)) |
| 56 | +``` |
| 57 | + |
| 58 | +```Python |
| 59 | +# Prim's algorithm |
| 60 | +class Solution: |
| 61 | + def getMinRiskValue(self, n, m, x, y, w): |
| 62 | + |
| 63 | + # construct graph |
| 64 | + adj = collections.defaultdict(list) |
| 65 | + for i in range(m): |
| 66 | + adj[x[i]].append((y[i], w[i])) |
| 67 | + adj[y[i]].append((x[i], w[i])) |
| 68 | + |
| 69 | + # Prim's algorithm with min heap |
| 70 | + mst = collections.defaultdict(list) |
| 71 | + min_heap = [(r, 0, v) for v, r in adj[0]] |
| 72 | + heapq.heapify(min_heap) |
| 73 | + |
| 74 | + while n not in mst: |
| 75 | + r, u, v = heapq.heappop(min_heap) |
| 76 | + if v not in mst: |
| 77 | + mst[u].append((v, r)) |
| 78 | + mst[v].append((u, r)) |
| 79 | + for nei, w in adj[v]: |
| 80 | + if nei not in mst: |
| 81 | + heapq.heappush(min_heap, (w, v, nei)) |
| 82 | + |
| 83 | + # dfs to search route from 0 to n |
| 84 | + dfs = [(0, None, float('-inf'))] |
| 85 | + while dfs: |
| 86 | + v, p, max_risk = dfs.pop() |
| 87 | + for a, r in mst[v]: |
| 88 | + cur_max = max(max_risk, r) |
| 89 | + if a == n: |
| 90 | + return cur_max |
| 91 | + if a != p: |
| 92 | + dfs.append((a, v, cur_max)) |
| 93 | +``` |
| 94 | + |
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