|
80 | 80 |
|
81 | 81 | <!-- 这里可写通用的实现逻辑 --> |
82 | 82 |
|
| 83 | +**方法一:动态规划** |
| 84 | + |
| 85 | +我们先根据给定的道路构建一个邻接表 $g,ドル其中 $g[i]$ 表示与城市 $i$ 直接相连的城市列表。 |
| 86 | + |
| 87 | +然后我们定义 $f[i][j]$ 表示 $targetPath$ 的第 $i$ 个城市与 $names$ 的第 $j$ 个城市匹配时,前 $i$ 个城市的最小编辑距离。 |
| 88 | + |
| 89 | +那么我们可以得到状态转移方程: |
| 90 | + |
| 91 | +$$ |
| 92 | +f[i][j] = \min_{k \in g[j]} f[i - 1][k] + (targetPath[i] \neq names[j]) |
| 93 | +$$ |
| 94 | + |
| 95 | +在状态转移的过程中,我们记录下每个状态的前驱城市,最后根据前驱城市数组 $pre$ 从后往前还原出最优路径。 |
| 96 | + |
| 97 | +时间复杂度 $O(m \times n^2),ドル空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是 $targetPath$ 和 $names$ 的长度。 |
| 98 | + |
83 | 99 | <!-- tabs:start --> |
84 | 100 |
|
85 | 101 | ### **Python3** |
86 | 102 |
|
87 | 103 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
88 | 104 |
|
89 | 105 | ```python |
90 | | - |
| 106 | +class Solution: |
| 107 | + def mostSimilar( |
| 108 | + self, n: int, roads: List[List[int]], names: List[str], targetPath: List[str] |
| 109 | + ) -> List[int]: |
| 110 | + g = [[] for _ in range(n)] |
| 111 | + for a, b in roads: |
| 112 | + g[a].append(b) |
| 113 | + g[b].append(a) |
| 114 | + m = len(targetPath) |
| 115 | + f = [[inf] * n for _ in range(m)] |
| 116 | + pre = [[-1] * n for _ in range(m)] |
| 117 | + for j, s in enumerate(names): |
| 118 | + f[0][j] = targetPath[0] != s |
| 119 | + for i in range(1, m): |
| 120 | + for j in range(n): |
| 121 | + for k in g[j]: |
| 122 | + if (t := f[i - 1][k] + (targetPath[i] != names[j])) < f[i][j]: |
| 123 | + f[i][j] = t |
| 124 | + pre[i][j] = k |
| 125 | + k = 0 |
| 126 | + mi = inf |
| 127 | + for j in range(n): |
| 128 | + if f[-1][j] < mi: |
| 129 | + mi = f[-1][j] |
| 130 | + k = j |
| 131 | + ans = [0] * m |
| 132 | + for i in range(m - 1, -1, -1): |
| 133 | + ans[i] = k |
| 134 | + k = pre[i][k] |
| 135 | + return ans |
91 | 136 | ``` |
92 | 137 |
|
93 | 138 | ### **Java** |
94 | 139 |
|
95 | 140 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
96 | 141 |
|
97 | 142 | ```java |
| 143 | +class Solution { |
| 144 | + public List<Integer> mostSimilar(int n, int[][] roads, String[] names, String[] targetPath) { |
| 145 | + List<Integer>[] g = new List[n]; |
| 146 | + Arrays.setAll(g, i -> new ArrayList<>()); |
| 147 | + for (int[] r : roads) { |
| 148 | + int a = r[0], b = r[1]; |
| 149 | + g[a].add(b); |
| 150 | + g[b].add(a); |
| 151 | + } |
| 152 | + int m = targetPath.length; |
| 153 | + final int inf = 1 << 30; |
| 154 | + int[][] f = new int[m][n]; |
| 155 | + int[][] pre = new int[m][n]; |
| 156 | + for (int i = 0; i < m; i++) { |
| 157 | + Arrays.fill(f[i], inf); |
| 158 | + Arrays.fill(pre[i], -1); |
| 159 | + } |
| 160 | + for (int j = 0; j < n; ++j) { |
| 161 | + f[0][j] = targetPath[0].equals(names[j]) ? 0 : 1; |
| 162 | + } |
| 163 | + for (int i = 1; i < m; ++i) { |
| 164 | + for (int j = 0; j < n; ++j) { |
| 165 | + for (int k : g[j]) { |
| 166 | + int t = f[i - 1][k] + (targetPath[i].equals(names[j]) ? 0 : 1); |
| 167 | + if (t < f[i][j]) { |
| 168 | + f[i][j] = t; |
| 169 | + pre[i][j] = k; |
| 170 | + } |
| 171 | + } |
| 172 | + } |
| 173 | + } |
| 174 | + int mi = inf, k = 0; |
| 175 | + for (int j = 0; j < n; ++j) { |
| 176 | + if (f[m - 1][j] < mi) { |
| 177 | + mi = f[m - 1][j]; |
| 178 | + k = j; |
| 179 | + } |
| 180 | + } |
| 181 | + List<Integer> ans = new ArrayList<>(); |
| 182 | + for (int i = m - 1; i >= 0; --i) { |
| 183 | + ans.add(k); |
| 184 | + k = pre[i][k]; |
| 185 | + } |
| 186 | + Collections.reverse(ans); |
| 187 | + return ans; |
| 188 | + } |
| 189 | +} |
| 190 | +``` |
| 191 | + |
| 192 | +### **C++** |
| 193 | + |
| 194 | +```cpp |
| 195 | +class Solution { |
| 196 | +public: |
| 197 | + vector<int> mostSimilar(int n, vector<vector<int>>& roads, vector<string>& names, vector<string>& targetPath) { |
| 198 | + vector<int> g[n]; |
| 199 | + for (auto& r : roads) { |
| 200 | + int a = r[0], b = r[1]; |
| 201 | + g[a].push_back(b); |
| 202 | + g[b].push_back(a); |
| 203 | + } |
| 204 | + int m = targetPath.size(); |
| 205 | + int f[m][n]; |
| 206 | + int pre[m][n]; |
| 207 | + memset(f, 0x3f, sizeof(f)); |
| 208 | + memset(pre, -1, sizeof(pre)); |
| 209 | + for (int j = 0; j < n; ++j) { |
| 210 | + f[0][j] = targetPath[0] != names[j]; |
| 211 | + } |
| 212 | + for (int i = 1; i < m; ++i) { |
| 213 | + for (int j = 0; j < n; ++j) { |
| 214 | + for (int k : g[j]) { |
| 215 | + int t = f[i - 1][k] + (targetPath[i] != names[j]); |
| 216 | + if (t < f[i][j]) { |
| 217 | + f[i][j] = t; |
| 218 | + pre[i][j] = k; |
| 219 | + } |
| 220 | + } |
| 221 | + } |
| 222 | + } |
| 223 | + int k = 0; |
| 224 | + int mi = 1 << 30; |
| 225 | + for (int j = 0; j < n; ++j) { |
| 226 | + if (f[m - 1][j] < mi) { |
| 227 | + mi = f[m - 1][j]; |
| 228 | + k = j; |
| 229 | + } |
| 230 | + } |
| 231 | + vector<int> ans(m); |
| 232 | + for (int i = m - 1; ~i; --i) { |
| 233 | + ans[i] = k; |
| 234 | + k = pre[i][k]; |
| 235 | + } |
| 236 | + return ans; |
| 237 | + } |
| 238 | +}; |
| 239 | +``` |
| 240 | + |
| 241 | +### **Go** |
| 242 | + |
| 243 | +```go |
| 244 | +func mostSimilar(n int, roads [][]int, names []string, targetPath []string) []int { |
| 245 | + g := make([][]int, n) |
| 246 | + for _, r := range roads { |
| 247 | + a, b := r[0], r[1] |
| 248 | + g[a] = append(g[a], b) |
| 249 | + g[b] = append(g[b], a) |
| 250 | + } |
| 251 | + m := len(targetPath) |
| 252 | + const inf = 1 << 30 |
| 253 | + f := make([][]int, m) |
| 254 | + pre := make([][]int, m) |
| 255 | + for i := range f { |
| 256 | + f[i] = make([]int, n) |
| 257 | + pre[i] = make([]int, n) |
| 258 | + for j := range f[i] { |
| 259 | + f[i][j] = inf |
| 260 | + pre[i][j] = -1 |
| 261 | + } |
| 262 | + } |
| 263 | + for j, s := range names { |
| 264 | + if targetPath[0] != s { |
| 265 | + f[0][j] = 1 |
| 266 | + } else { |
| 267 | + f[0][j] = 0 |
| 268 | + } |
| 269 | + } |
| 270 | + for i := 1; i < m; i++ { |
| 271 | + for j := 0; j < n; j++ { |
| 272 | + for _, k := range g[j] { |
| 273 | + t := f[i-1][k] |
| 274 | + if targetPath[i] != names[j] { |
| 275 | + t++ |
| 276 | + } |
| 277 | + if t < f[i][j] { |
| 278 | + f[i][j] = t |
| 279 | + pre[i][j] = k |
| 280 | + } |
| 281 | + } |
| 282 | + } |
| 283 | + } |
| 284 | + mi, k := inf, 0 |
| 285 | + for j := 0; j < n; j++ { |
| 286 | + if f[m-1][j] < mi { |
| 287 | + mi = f[m-1][j] |
| 288 | + k = j |
| 289 | + } |
| 290 | + } |
| 291 | + ans := make([]int, m) |
| 292 | + for i := m - 1; i >= 0; i-- { |
| 293 | + ans[i] = k |
| 294 | + k = pre[i][k] |
| 295 | + } |
| 296 | + return ans |
| 297 | +} |
| 298 | +``` |
98 | 299 |
|
| 300 | +### **TypeScript** |
| 301 | + |
| 302 | +```ts |
| 303 | +function mostSimilar( |
| 304 | + n: number, |
| 305 | + roads: number[][], |
| 306 | + names: string[], |
| 307 | + targetPath: string[], |
| 308 | +): number[] { |
| 309 | + const g: number[][] = Array.from({ length: n }, () => []); |
| 310 | + for (const [a, b] of roads) { |
| 311 | + g[a].push(b); |
| 312 | + g[b].push(a); |
| 313 | + } |
| 314 | + const m = targetPath.length; |
| 315 | + const f = Array.from({ length: m }, () => Array.from({ length: n }, () => Infinity)); |
| 316 | + const pre: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => -1)); |
| 317 | + for (let j = 0; j < n; ++j) { |
| 318 | + f[0][j] = names[j] === targetPath[0] ? 0 : 1; |
| 319 | + } |
| 320 | + for (let i = 1; i < m; ++i) { |
| 321 | + for (let j = 0; j < n; ++j) { |
| 322 | + for (const k of g[j]) { |
| 323 | + const t = f[i - 1][k] + (names[j] === targetPath[i] ? 0 : 1); |
| 324 | + if (t < f[i][j]) { |
| 325 | + f[i][j] = t; |
| 326 | + pre[i][j] = k; |
| 327 | + } |
| 328 | + } |
| 329 | + } |
| 330 | + } |
| 331 | + let k = 0; |
| 332 | + let mi = Infinity; |
| 333 | + for (let j = 0; j < n; ++j) { |
| 334 | + if (f[m - 1][j] < mi) { |
| 335 | + mi = f[m - 1][j]; |
| 336 | + k = j; |
| 337 | + } |
| 338 | + } |
| 339 | + const ans: number[] = Array(m).fill(0); |
| 340 | + for (let i = m - 1; ~i; --i) { |
| 341 | + ans[i] = k; |
| 342 | + k = pre[i][k]; |
| 343 | + } |
| 344 | + return ans; |
| 345 | +} |
99 | 346 | ``` |
100 | 347 |
|
101 | 348 | ### **...** |
|
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