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.
\n\nhouses = [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.
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Example 1:
\n\n\nInput: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3\nOutput: 9\nExplanation: Paint houses of this way [1,2,2,1,1]\nThis array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].\nCost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.\n\n\nExample 2:
\n\n\nInput: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3\nOutput: 11\nExplanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2]\nThis array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. \nCost of paint the first and last house (10 + 1) = 11.\n\n\nExample 3:
\n\n\nInput: houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3\nOutput: -1\nExplanation: Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.\n\n\n\n
Constraints:
\n\nm == houses.length == cost.lengthn == cost[i].length1 <= m <= 1001 <= n <= 201 <= target <= m0 <= houses[i] <= n1 <= cost[i][j] <= 104在一个小城市里，有 m 个房子排成一排，你需要给每个房子涂上 n 种颜色之一（颜色编号为 1 到 n ）。有的房子去年夏天已经涂过颜色了，所以这些房子不可以被重新涂色。
我们将连续相同颜色尽可能多的房子称为一个街区。（比方说 houses = [1,2,2,3,3,2,1,1] ，它包含 5 个街区 [{1}, {2,2}, {3,3}, {2}, {1,1}] 。）
给你一个数组 houses ，一个 m * n 的矩阵 cost 和一个整数 target ，其中：
houses[i]：是第 i 个房子的颜色，0 表示这个房子还没有被涂色。cost[i][j]：是将第 i 个房子涂成颜色 j+1 的花费。请你返回房子涂色方案的最小总花费，使得每个房子都被涂色后，恰好组成 target 个街区。如果没有可用的涂色方案，请返回 -1 。
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示例 1：
\n\n\n输入：houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3\n输出：9\n解释：房子涂色方案为 [1,2,2,1,1]\n此方案包含 target = 3 个街区，分别是 [{1}, {2,2}, {1,1}]。\n涂色的总花费为 (1 + 1 + 1 + 1 + 5) = 9。\n\n\n示例 2：
\n\n\n输入：houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3\n输出：11\n解释：有的房子已经被涂色了，在此基础上涂色方案为 [2,2,1,2,2]\n此方案包含 target = 3 个街区，分别是 [{2,2}, {1}, {2,2}]。\n给第一个和最后一个房子涂色的花费为 (10 + 1) = 11。\n\n\n示例 3：
\n\n\n输入：houses = [0,0,0,0,0], cost = [[1,10],[10,1],[1,10],[10,1],[1,10]], m = 5, n = 2, target = 5\n输出：5\n\n\n
示例 4：
\n\n\n输入：houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3\n输出：-1\n解释：房子已经被涂色并组成了 4 个街区，分别是 [{3},{1},{2},{3}] ，无法形成 target = 3 个街区。\n\n\n\n\n
提示：
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