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feat: add solutions to lc problem: No.1473
No.1473.Paint House III
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Diff for: ‎solution/1400-1499/1473.Paint House III/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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我们定义 $f[i][j][k]$ 表示将下标 $[0,..i]$ 的房子涂上颜色，最后一个房子的颜色为 $j$，且恰好形成 $k$ 个街区的最小花费。那么答案就是 $f[m-1][j][target]$，其中 $j$ 的取值范围为 $[1,..n]$。初始时，我们判断下标为 $0$ 的房子是否已经涂色，如果未涂色，那么 $f[0][j][1] = cost[0][j - 1]$，其中 $j \in [1,..n]$。如果已经涂色，那么 $f[0][houses[0]][1] = 0$。其他的 $f[i][j][k]$ 的值都初始化为 $\infty$。
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接下来，我们从下标 $i=1$ 开始遍历，对于每个 $i$，我们判断下标为 $i$ 的房子是否已经涂色：
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如果未涂色，那么我们可以将下标为 $i$ 的房子涂成颜色 $j$，我们枚举街区的数量 $k$，其中 $k \in [1,..min(target, i + 1)]$，并且枚举下标为 $i$ 的房子的前一个房子的颜色 $j_0$，其中 $j_0 \in [1,..n]$，那么我们可以得到状态转移方程：
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$$
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f[i][j][k] = \min_{j_0 \in [1,..n]} \{ f[i - 1][j_0][k - (j \neq j_0)] + cost[i][j - 1] \}
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$$
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如果已经涂色，那么我们可以将下标为 $i$ 的房子涂成颜色 $j$，我们枚举街区的数量 $k$，其中 $k \in [1,..min(target, i + 1)]$，并且枚举下标为 $i$ 的房子的前一个房子的颜色 $j_0$，其中 $j_0 \in [1,..n]$，那么我们可以得到状态转移方程：
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$$
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f[i][j][k] = \min_{j_0 \in [1,..n]} \{ f[i - 1][j_0][k - (j \neq j_0)] \}
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$$
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最后，我们返回 $f[m - 1][j][target]$，其中 $j \in [1,..n]$，如果所有的 $f[m - 1][j][target]$ 的值都为 $\infty$，那么返回 $-1$。
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时间复杂度 $O(m \times n^2 \times target)$，空间复杂度 $O(m \times n \times target)$。其中 $m$, $n$, $target$ 分别为房子的数量，颜色的数量，街区的数量。
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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 minCost(self, houses: List[int], cost: List[List[int]], m: int, n: int, target: int) -> int:
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f = [[[inf] * (target + 1) for _ in range(n + 1)] for _ in range(m)]
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if houses[0] == 0:
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for j, c in enumerate(cost[0], 1):
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f[0][j][1] = c
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else:
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f[0][houses[0]][1] = 0
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for i in range(1, m):
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if houses[i] == 0:
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for j in range(1, n + 1):
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for k in range(1, min(target + 1, i + 2)):
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for j0 in range(1, n + 1):
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if j == j0:
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f[i][j][k] = min(
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f[i][j][k], f[i - 1][j][k] + cost[i][j - 1])
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else:
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f[i][j][k] = min(
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f[i][j][k], f[i - 1][j0][k - 1] + cost[i][j - 1])
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else:
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j = houses[i]
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for k in range(1, min(target + 1, i + 2)):
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for j0 in range(1, n + 1):
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if j == j0:
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f[i][j][k] = min(f[i][j][k], f[i - 1][j][k])
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else:
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f[i][j][k] = min(f[i][j][k], f[i - 1][j0][k - 1])
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ans = min(f[-1][j][target] for j in range(1, n + 1))
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return -1 if ans >= inf else ans
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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 minCost(int[] houses, int[][] cost, int m, int n, int target) {
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int[][][] f = new int[m][n + 1][target + 1];
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final int inf = 1 << 30;
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for (int[][] g : f) {
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for (int[] e : g) {
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Arrays.fill(e, inf);
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}
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}
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if (houses[0] == 0) {
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for (int j = 1; j <= n; ++j) {
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f[0][j][1] = cost[0][j - 1];
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}
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} else {
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f[0][houses[0]][1] = 0;
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}
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for (int i = 1; i < m; ++i) {
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if (houses[i] == 0) {
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for (int j = 1; j <= n; ++j) {
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for (int k = 1; k <= Math.min(target, i + 1); ++k) {
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for (int j0 = 1; j0 <= n; ++j0) {
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if (j == j0) {
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f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j][k] + cost[i][j - 1]);
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} else {
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f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j0][k - 1] + cost[i][j - 1]);
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}
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}
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}
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}
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} else {
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int j = houses[i];
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for (int k = 1; k <= Math.min(target, i + 1); ++k) {
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for (int j0 = 1; j0 <= n; ++j0) {
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if (j == j0) {
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f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j][k]);
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} else {
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f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j0][k - 1]);
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}
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}
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}
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}
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}
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int ans = inf;
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for (int j = 1; j <= n; ++j) {
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ans = Math.min(ans, f[m - 1][j][target]);
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}
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return ans >= inf ? -1 : ans;
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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 minCost(vector<int>& houses, vector<vector<int>>& cost, int m, int n, int target) {
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int f[m][n + 1][target + 1];
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memset(f, 0x3f, sizeof(f));
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if (houses[0] == 0) {
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for (int j = 1; j <= n; ++j) {
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f[0][j][1] = cost[0][j - 1];
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}
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} else {
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f[0][houses[0]][1] = 0;
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}
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for (int i = 1; i < m; ++i) {
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if (houses[i] == 0) {
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for (int j = 1; j <= n; ++j) {
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for (int k = 1; k <= min(target, i + 1); ++k) {
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for (int j0 = 1; j0 <= n; ++j0) {
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if (j == j0) {
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f[i][j][k] = min(f[i][j][k], f[i - 1][j][k] + cost[i][j - 1]);
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} else {
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f[i][j][k] = min(f[i][j][k], f[i - 1][j0][k - 1] + cost[i][j - 1]);
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}
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}
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}
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}
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} else {
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int j = houses[i];
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for (int k = 1; k <= min(target, i + 1); ++k) {
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for (int j0 = 1; j0 <= n; ++j0) {
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if (j == j0) {
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f[i][j][k] = min(f[i][j][k], f[i - 1][j][k]);
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} else {
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f[i][j][k] = min(f[i][j][k], f[i - 1][j0][k - 1]);
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}
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}
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}
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}
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}
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int ans = 0x3f3f3f3f;
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for (int j = 1; j <= n; ++j) {
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ans = min(ans, f[m - 1][j][target]);
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}
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return ans == 0x3f3f3f3f ? -1 : ans;
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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 minCost(houses []int, cost [][]int, m int, n int, target int) int {
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f := make([][][]int, m)
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const inf = 1 << 30
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for i := range f {
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f[i] = make([][]int, n+1)
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for j := range f[i] {
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f[i][j] = make([]int, target+1)
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for k := range f[i][j] {
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f[i][j][k] = inf
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}
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}
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}
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if houses[0] == 0 {
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for j := 1; j <= n; j++ {
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f[0][j][1] = cost[0][j-1]
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}
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} else {
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f[0][houses[0]][1] = 0
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}
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for i := 1; i < m; i++ {
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if houses[i] == 0 {
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for j := 1; j <= n; j++ {
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for k := 1; k <= target && k <= i+1; k++ {
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for j0 := 1; j0 <= n; j0++ {
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if j == j0 {
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f[i][j][k] = min(f[i][j][k], f[i-1][j][k]+cost[i][j-1])
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} else {
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f[i][j][k] = min(f[i][j][k], f[i-1][j0][k-1]+cost[i][j-1])
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}
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}
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}
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}
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} else {
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j := houses[i]
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for k := 1; k <= target && k <= i+1; k++ {
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for j0 := 1; j0 <= n; j0++ {
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if j == j0 {
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f[i][j][k] = min(f[i][j][k], f[i-1][j][k])
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} else {
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f[i][j][k] = min(f[i][j][k], f[i-1][j0][k-1])
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}
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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 j := 1; j <= n; j++ {
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ans = min(ans, f[m-1][j][target])
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}
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if ans == inf {
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return -1
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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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```
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### **...**

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