|
46 | 46 | <li><code>conections[i][0] != conections[i][1]</code></li> |
47 | 47 | </ol> |
48 | 48 |
|
49 | | - |
50 | 49 | ## 解法 |
51 | 50 |
|
52 | 51 | <!-- 这里可写通用的实现逻辑 --> |
53 | 52 |
|
| 53 | +最小生成树 + 并查集。 |
| 54 | + |
| 55 | +并查集模板: |
| 56 | + |
| 57 | +模板 1——朴素并查集: |
| 58 | + |
| 59 | +```python |
| 60 | +# 初始化,p存储每个点的父节点 |
| 61 | +p = list(range(n)) |
| 62 | + |
| 63 | +# 返回x的祖宗节点 |
| 64 | +def find(x): |
| 65 | + if p[x] != x: |
| 66 | + # 路径压缩 |
| 67 | + p[x] = find(p[x]) |
| 68 | + return p[x] |
| 69 | + |
| 70 | +# 合并a和b所在的两个集合 |
| 71 | +p[find(a)] = find(b) |
| 72 | +``` |
| 73 | + |
| 74 | +模板 2——维护 size 的并查集: |
| 75 | + |
| 76 | +```python |
| 77 | +# 初始化,p存储每个点的父节点,size只有当节点是祖宗节点时才有意义,表示祖宗节点所在集合中,点的数量 |
| 78 | +p = list(range(n)) |
| 79 | +size = [1] * n |
| 80 | + |
| 81 | +# 返回x的祖宗节点 |
| 82 | +def find(x): |
| 83 | + if p[x] != x: |
| 84 | + # 路径压缩 |
| 85 | + p[x] = find(p[x]) |
| 86 | + return p[x] |
| 87 | + |
| 88 | +# 合并a和b所在的两个集合 |
| 89 | +if find(a) != find(b): |
| 90 | + size[find(b)] += size[find(a)] |
| 91 | + p[find(a)] = find(b) |
| 92 | +``` |
| 93 | + |
| 94 | +模板 3——维护到祖宗节点距离的并查集: |
| 95 | + |
| 96 | +```python |
| 97 | +# 初始化,p存储每个点的父节点,d[x]存储x到p[x]的距离 |
| 98 | +p = list(range(n)) |
| 99 | +d = [0] * n |
| 100 | + |
| 101 | +# 返回x的祖宗节点 |
| 102 | +def find(x): |
| 103 | + if p[x] != x: |
| 104 | + t = find(p[x]) |
| 105 | + d[x] += d[p[x]] |
| 106 | + p[x] = t |
| 107 | + return p[x] |
| 108 | + |
| 109 | +# 合并a和b所在的两个集合 |
| 110 | +p[find(a)] = find(b) |
| 111 | +d[find(a)] = distance |
| 112 | +``` |
| 113 | + |
54 | 114 | <!-- tabs:start --> |
55 | 115 |
|
56 | 116 | ### **Python3** |
57 | 117 |
|
58 | 118 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
59 | 119 |
|
60 | 120 | ```python |
61 | | - |
| 121 | +class Solution: |
| 122 | + def minimumCost(self, n: int, connections: List[List[int]]) -> int: |
| 123 | + p = list(range(n)) |
| 124 | + connections.sort(key=lambda x: x[2]) |
| 125 | + res = 0 |
| 126 | + |
| 127 | + def find(x): |
| 128 | + if p[x] != x: |
| 129 | + p[x] = find(p[x]) |
| 130 | + return p[x] |
| 131 | + |
| 132 | + def union(a, b): |
| 133 | + pa, pb = find(a - 1), find(b - 1) |
| 134 | + if pa == pb: |
| 135 | + return False |
| 136 | + p[pa] = pb |
| 137 | + return True |
| 138 | + |
| 139 | + for c1, c2, cost in connections: |
| 140 | + if union(c1, c2): |
| 141 | + n -= 1 |
| 142 | + res += cost |
| 143 | + if n == 1: |
| 144 | + return res |
| 145 | + return -1 |
62 | 146 | ``` |
63 | 147 |
|
64 | 148 | ### **Java** |
65 | 149 |
|
66 | 150 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
67 | 151 |
|
68 | 152 | ```java |
| 153 | +class Solution { |
| 154 | + private int[] p; |
| 155 | + |
| 156 | + public int minimumCost(int n, int[][] connections) { |
| 157 | + p = new int[n]; |
| 158 | + for (int i = 0; i < n; ++i) { |
| 159 | + p[i] = i; |
| 160 | + } |
| 161 | + Arrays.sort(connections, (a, b) -> a[2] - b[2]); |
| 162 | + int res = 0; |
| 163 | + for (int[] e : connections) { |
| 164 | + if (union(e[0], e[1])) { |
| 165 | + res += e[2]; |
| 166 | + --n; |
| 167 | + if (n == 1) { |
| 168 | + return res; |
| 169 | + } |
| 170 | + } |
| 171 | + } |
| 172 | + return -1; |
| 173 | + } |
| 174 | + |
| 175 | + private int find(int x) { |
| 176 | + if (p[x] != x) { |
| 177 | + p[x] = find(p[x]); |
| 178 | + } |
| 179 | + return p[x]; |
| 180 | + } |
| 181 | + |
| 182 | + private boolean union(int a, int b) { |
| 183 | + int pa = find(a - 1), pb = find(b - 1); |
| 184 | + if (pa == pb) { |
| 185 | + return false; |
| 186 | + } |
| 187 | + p[pa] = pb; |
| 188 | + return true; |
| 189 | + } |
| 190 | +} |
| 191 | +``` |
| 192 | + |
| 193 | +### **C++** |
| 194 | + |
| 195 | +```cpp |
| 196 | +class Solution { |
| 197 | +public: |
| 198 | + vector<int> p; |
| 199 | + |
| 200 | + int minimumCost(int n, vector<vector<int>> &connections) { |
| 201 | + p.resize(n); |
| 202 | + for (int i = 0; i < n; ++i) p[i] = i; |
| 203 | + auto cmp = [](auto &a, auto &b) |
| 204 | + { |
| 205 | + return a[2] < b[2]; |
| 206 | + }; |
| 207 | + sort(connections.begin(), connections.end(), cmp); |
| 208 | + int res = 0; |
| 209 | + for (auto e : connections) |
| 210 | + { |
| 211 | + if (unite(e[0], e[1])) |
| 212 | + { |
| 213 | + res += e[2]; |
| 214 | + --n; |
| 215 | + if (n == 1) return res; |
| 216 | + } |
| 217 | + } |
| 218 | + return -1; |
| 219 | + } |
| 220 | + |
| 221 | + int find(int x) { |
| 222 | + if (p[x] != x) p[x] = find(p[x]); |
| 223 | + return p[x]; |
| 224 | + } |
| 225 | + |
| 226 | + bool unite(int a, int b) { |
| 227 | + int pa = find(a - 1), pb = find(b - 1); |
| 228 | + if (pa == pb) return false; |
| 229 | + p[pa] = pb; |
| 230 | + return true; |
| 231 | + } |
| 232 | +}; |
| 233 | +``` |
69 | 234 | |
| 235 | +### **Go** |
| 236 | + |
| 237 | +```go |
| 238 | +var p []int |
| 239 | + |
| 240 | +func minimumCost(n int, connections [][]int) int { |
| 241 | + p = make([]int, n) |
| 242 | + for i := 0; i < len(p); i++ { |
| 243 | + p[i] = i |
| 244 | + } |
| 245 | + sort.Slice(connections, func(i, j int) bool { |
| 246 | + return connections[i][2] < connections[j][2] |
| 247 | + }) |
| 248 | + res := 0 |
| 249 | + for _, e := range connections { |
| 250 | + if union(e[0], e[1]) { |
| 251 | + res += e[2] |
| 252 | + n-- |
| 253 | + if n == 1 { |
| 254 | + return res |
| 255 | + } |
| 256 | + } |
| 257 | + } |
| 258 | + return -1 |
| 259 | +} |
| 260 | + |
| 261 | +func find(x int) int { |
| 262 | + if p[x] != x { |
| 263 | + p[x] = find(p[x]) |
| 264 | + } |
| 265 | + return p[x] |
| 266 | +} |
| 267 | + |
| 268 | +func union(a, b int) bool { |
| 269 | + pa, pb := find(a-1), find(b-1) |
| 270 | + if pa == pb { |
| 271 | + return false |
| 272 | + } |
| 273 | + p[pa] = pb |
| 274 | + return true |
| 275 | +} |
70 | 276 | ``` |
71 | 277 |
|
72 | 278 | ### **...** |
|
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