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feat: add solutions to lc problem: No.1066
No.1066.Campus Bikes II
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‎solution/1000-1099/1066.Campus Bikes II/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：状态压缩动态规划**
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我们定义 $f[i][j]$ 表示前 $i$ 个工人分配到自行车的状态为 $j$ 时的最小曼哈顿距离总和，其中 $j$ 是一个二进制数，表示自行车的分配情况。初始时 $f[0][0]=0$，其余 $f[0][j]=+\infty$。
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考虑 $f[i][j]$，我们枚举第 $i$ 个工人分配到的自行车的编号 $k$，那么 $f[i][j]$ 可以从 $f[i-1][j\oplus 2^k]$ 转移而来，其中 $\oplus$ 表示异或运算。这是因为 $f[i-1][j\oplus 2^k]$ 表示前 $i-1$ 个工人分配到自行车的状态为 $j\oplus 2^k$ 时的最小曼哈顿距离总和，而第 $i$ 个工人分配到自行车 $k$ 时，其曼哈顿距离为 $|worker[i]-bike[k]|$，其中 $|x|$ 表示 $x$ 的绝对值。因此我们可以列出状态转移方程：
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$$
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f[i][j]=\min_{k=0}^{m-1}\{f[i-1][j\oplus 2^k]+|worker[i]-bike[k]|\}
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$$
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最终的答案为 $\min_{j=0}^{2^m-1}\{f[n][j]\}$。
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时间复杂度 $O(n \times 2^m \times m)$，空间复杂度 $O(n \times 2^m)$。其中 $n$ 和 $m$ 分别是工人和自行车的数量。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def assignBikes(self, workers: List[List[int]], bikes: List[List[int]]) -> int:
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n, m = len(workers), len(bikes)
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f = [[inf] * (1 << m) for _ in range(n + 1)]
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f[0][0] = 0
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for i, (x1, y1) in enumerate(workers, 1):
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for j in range(1 << m):
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for k, (x2, y2) in enumerate(bikes):
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if j >> k & 1:
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f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + abs(x1 - x2) + abs(y1 - y2))
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return min(f[n])
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int assignBikes(int[][] workers, int[][] bikes) {
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int n = workers.length;
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int m = bikes.length;
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int[][] f = new int[n + 1][1 << m];
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for (var g : f) {
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Arrays.fill(g, 1 << 30);
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}
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f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j < 1 << m; ++j) {
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for (int k = 0; k < m; ++k) {
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if ((j >> k & 1) == 1) {
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int d = Math.abs(workers[i - 1][0] - bikes[k][0]) + Math.abs(workers[i - 1][1] - bikes[k][1]);
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f[i][j] = Math.min(f[i][j], f[i - 1][j ^ (1 << k)] + d);
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}
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}
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}
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}
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return Arrays.stream(f[n]).min().getAsInt();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) {
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int n = workers.size(), m = bikes.size();
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int f[n + 1][1 << m];
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memset(f, 0x3f, sizeof(f));
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f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j < 1 << m; ++j) {
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for (int k = 0; k < m; ++k) {
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if (j >> k & 1) {
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int d = abs(workers[i - 1][0] - bikes[k][0]) + abs(workers[i - 1][1] - bikes[k][1]);
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f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + d);
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}
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}
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}
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}
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return *min_element(f[n], f[n] + (1 << m));
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}
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};
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```
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### **Go**
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```go
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func assignBikes(workers [][]int, bikes [][]int) int {
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n, m := len(workers), len(bikes)
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f := make([][]int, n+1)
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const inf = 1 << 30
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for i := range f {
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f[i] = make([]int, 1<<m)
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for j := range f[i] {
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f[i][j] = inf
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}
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}
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f[0][0] = 0
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for i := 1; i <= n; i++ {
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for j := 0; j < 1<<m; j++ {
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for k := 0; k < m; k++ {
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if j>>k&1 == 1 {
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d := abs(workers[i-1][0]-bikes[k][0]) + abs(workers[i-1][1]-bikes[k][1])
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f[i][j] = min(f[i][j], f[i-1][j^(1<<k)]+d)
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}
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}
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}
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}
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ans := inf
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for _, x := range f[n] {
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ans = min(ans, x)
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **TypeScript**
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```ts
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function assignBikes(workers: number[][], bikes: number[][]): number {
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const n = workers.length;
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const m = bikes.length;
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const inf = 1 << 30;
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const f: number[][] = new Array(n + 1)
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.fill(0)
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.map(() => new Array(1 << m).fill(inf));
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f[0][0] = 0;
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for (let i = 1; i <= n; ++i) {
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for (let j = 0; j < 1 << m; ++j) {
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for (let k = 0; k < m; ++k) {
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if (((j >> k) & 1) === 1) {
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const d =
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Math.abs(workers[i - 1][0] - bikes[k][0]) +
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Math.abs(workers[i - 1][1] - bikes[k][1]);
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f[i][j] = Math.min(f[i][j], f[i - 1][j ^ (1 << k)] + d);
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}
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}
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}
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}
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return Math.min(...f[n]);
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}
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```
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### **...**

‎solution/1000-1099/1066.Campus Bikes II/README_EN.md

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@@ -58,13 +58,142 @@ We first assign bike 0 to worker 0, then assign bike 1 to worker 1 or worker 2,
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### **Python3**
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```python
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class Solution:
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def assignBikes(self, workers: List[List[int]], bikes: List[List[int]]) -> int:
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n, m = len(workers), len(bikes)
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f = [[inf] * (1 << m) for _ in range(n + 1)]
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f[0][0] = 0
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for i, (x1, y1) in enumerate(workers, 1):
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for j in range(1 << m):
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for k, (x2, y2) in enumerate(bikes):
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if j >> k & 1:
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f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + abs(x1 - x2) + abs(y1 - y2))
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return min(f[n])
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```
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### **Java**
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```java
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class Solution {
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public int assignBikes(int[][] workers, int[][] bikes) {
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int n = workers.length;
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int m = bikes.length;
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int[][] f = new int[n + 1][1 << m];
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for (var g : f) {
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Arrays.fill(g, 1 << 30);
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}
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f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j < 1 << m; ++j) {
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for (int k = 0; k < m; ++k) {
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if ((j >> k & 1) == 1) {
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int d = Math.abs(workers[i - 1][0] - bikes[k][0]) + Math.abs(workers[i - 1][1] - bikes[k][1]);
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f[i][j] = Math.min(f[i][j], f[i - 1][j ^ (1 << k)] + d);
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}
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}
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}
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}
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return Arrays.stream(f[n]).min().getAsInt();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) {
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int n = workers.size(), m = bikes.size();
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int f[n + 1][1 << m];
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memset(f, 0x3f, sizeof(f));
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f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j < 1 << m; ++j) {
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for (int k = 0; k < m; ++k) {
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if (j >> k & 1) {
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int d = abs(workers[i - 1][0] - bikes[k][0]) + abs(workers[i - 1][1] - bikes[k][1]);
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f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + d);
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}
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}
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}
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}
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return *min_element(f[n], f[n] + (1 << m));
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}
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};
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```
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### **Go**
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```go
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func assignBikes(workers [][]int, bikes [][]int) int {
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n, m := len(workers), len(bikes)
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f := make([][]int, n+1)
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const inf = 1 << 30
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for i := range f {
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f[i] = make([]int, 1<<m)
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for j := range f[i] {
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f[i][j] = inf
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}
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}
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f[0][0] = 0
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for i := 1; i <= n; i++ {
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for j := 0; j < 1<<m; j++ {
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for k := 0; k < m; k++ {
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if j>>k&1 == 1 {
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d := abs(workers[i-1][0]-bikes[k][0]) + abs(workers[i-1][1]-bikes[k][1])
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f[i][j] = min(f[i][j], f[i-1][j^(1<<k)]+d)
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}
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}
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}
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}
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ans := inf
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for _, x := range f[n] {
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ans = min(ans, x)
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **TypeScript**
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```ts
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function assignBikes(workers: number[][], bikes: number[][]): number {
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const n = workers.length;
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const m = bikes.length;
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const inf = 1 << 30;
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const f: number[][] = new Array(n + 1)
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.fill(0)
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.map(() => new Array(1 << m).fill(inf));
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f[0][0] = 0;
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for (let i = 1; i <= n; ++i) {
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for (let j = 0; j < 1 << m; ++j) {
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for (let k = 0; k < m; ++k) {
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if (((j >> k) & 1) === 1) {
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const d =
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Math.abs(workers[i - 1][0] - bikes[k][0]) +
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Math.abs(workers[i - 1][1] - bikes[k][1]);
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f[i][j] = Math.min(f[i][j], f[i - 1][j ^ (1 << k)] + d);
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}
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}
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}
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}
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return Math.min(...f[n]);
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}
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```
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### **...**

‎solution/1000-1099/1066.Campus Bikes II/Solution.cpp

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class Solution {
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public:
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int assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) {
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int n = workers.size(), m = bikes.size();
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int f[n + 1][1 << m];
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memset(f, 0x3f, sizeof(f));
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f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j < 1 << m; ++j) {
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for (int k = 0; k < m; ++k) {
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if (j >> k & 1) {
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int d = abs(workers[i - 1][0] - bikes[k][0]) + abs(workers[i - 1][1] - bikes[k][1]);
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f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + d);
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}
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}
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}
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}
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return *min_element(f[n], f[n] + (1 << m));
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}
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};

‎solution/1000-1099/1066.Campus Bikes II/Solution.go

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func assignBikes(workers [][]int, bikes [][]int) int {
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n, m := len(workers), len(bikes)
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f := make([][]int, n+1)
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const inf = 1 << 30
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for i := range f {
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f[i] = make([]int, 1<<m)
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for j := range f[i] {
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f[i][j] = inf
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}
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}
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f[0][0] = 0
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for i := 1; i <= n; i++ {
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for j := 0; j < 1<<m; j++ {
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for k := 0; k < m; k++ {
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if j>>k&1 == 1 {
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d := abs(workers[i-1][0]-bikes[k][0]) + abs(workers[i-1][1]-bikes[k][1])
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f[i][j] = min(f[i][j], f[i-1][j^(1<<k)]+d)
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}
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}
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}
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}
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ans := inf
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for _, x := range f[n] {
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ans = min(ans, x)
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}

‎solution/1000-1099/1066.Campus Bikes II/Solution.java

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class Solution {
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public int assignBikes(int[][] workers, int[][] bikes) {
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int n = workers.length;
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int m = bikes.length;
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int[][] f = new int[n + 1][1 << m];
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for (var g : f) {
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Arrays.fill(g, 1 << 30);
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}
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f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j < 1 << m; ++j) {
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for (int k = 0; k < m; ++k) {
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if ((j >> k & 1) == 1) {
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int d = Math.abs(workers[i - 1][0] - bikes[k][0]) + Math.abs(workers[i - 1][1] - bikes[k][1]);
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f[i][j] = Math.min(f[i][j], f[i - 1][j ^ (1 << k)] + d);
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}
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}
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}
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}
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return Arrays.stream(f[n]).min().getAsInt();
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}
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}

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