leetcode 1473. Paint House III（粉刷房子III）

There is a row of m houses in a small city, each house must be painted with one of the n colors (labeled from 1 to n), some houses that have been painted last summer should not be painted again.

A neighborhood is a maximal group of continuous houses that are painted with the same color.

For example: houses = [1,2,2,3,3,2,1,1] contains 5 neighborhoods [{1}, {2,2}, {3,3}, {2}, {1,1}].
Given an array houses, an m x n matrix cost and an integer target where:

houses[i]: is the color of the house i, and 0 if the house is not painted yet.
cost[i][j]: is the cost of paint the house i with the color j + 1.
Return the minimum cost of painting all the remaining houses in such a way that there are exactly target neighborhoods. If it is not possible, return -1.

Example 1:

Input: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 9
Explanation: Paint houses of this way [1,2,2,1,1]
This array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].
Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.
Example 2:

Input: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 11
Explanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2]
This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}].
Cost of paint the first and last house (10 + 1) = 11.

houses数组中的数字表示第i 个房子的颜色，一共m个房子。
把连续相同颜色的分为1组，那么houses = [1,2,2,3,3,2,1,1] 这种的就分为5组 [{1}, {2,2}, {3,3}, {2}, {1,1}].

现在要把houses分成target个组，就需要把houses刷上不同的颜色以达到目的，
cost中就有n个颜色，cost = [[1,10],[10,1],[10,1],[1,10],[5,1]]这个的意思就是给第1个房子刷颜色1需要的cost是1，刷颜色2的cost是10，
同样的，给第2个房子刷1的cost是10，刷2cost是1，cost[i][j]就是第i个房子刷j+1颜色的cost

如果第i 个house已经有颜色（house[i] != 0), 保持原来的颜色，不再刷新的颜色。是0才能刷新的颜色。

求能分为target个组需要的cost, 如果不能达到，返回-1.

思路：

DP
变量较多，属于高维度DP，比较有难度。

先理一下变量，总的来说有3个变量，m个房子，n种颜色，要分为target个组。
定义dp[target+1][m+1][n+1]

首先，m和n都是>=1的，那么在它们为0时是不需要刷颜色的，cost=0
所以dp[0][0][*] = 0.

然后来看状态转移：
dp[k][i][ci] 表示 把house分为k个组，用前 i 个房子，第i个房子的颜色为ci，需要的cost

现在从dp[k-1][i-1][c_i-1]出发，假如第 i 个颜色 ci != c_i-1，那么就会形成一个新的组，现在有i个房子
于是dp[k-1][i-1][c_i-1]就转变为dp[k][i][ci]
如果ci == c_i-1, 就不会形成新的组，dp[k][i-1][c_i-1] 会变成 dp[k][i][ci]
那还要加上刷ci颜色的cost, 如果house已经有颜色了，不用刷，cost=0,
反之，house没有颜色，需要刷，则cost = cost[i][ci]
总结一下，状态转移为：
dp[k][i][ci] = min(dp[k][i][ci], dp[k - (ci != c_i-1)][i-1][c_i-1] + cost)

最后要返回的是最后一个房子在所有颜色中cost最小的那个，也就是min(dp[target][m])

可能还是比较难理解，在代码中再次给出注释

代码中要注意下标，题目中的房子，颜色都是从1开始的，而数组下标是从0开始。

    public int minCost(int[] houses, int[][