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

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

<ul>
	<li>For example: <code>houses = [1,2,2,3,3,2,1,1]</code> contains <code>5</code> neighborhoods <code>[{1}, {2,2}, {3,3}, {2}, {1,1}]</code>.</li>
</ul>

<p>Given an array <code>houses</code>, an <code>m x n</code> matrix <code>cost</code> and an integer <code>target</code> where:</p>

<ul>
	<li><code>houses[i]</code>: is the color of the house <code>i</code>, and <code>0</code> if the house is not painted yet.</li>
	<li><code>cost[i][j]</code>: is the cost of paint the house <code>i</code> with the color <code>j + 1</code>.</li>
</ul>

<p>Return <em>the minimum cost of painting all the remaining houses in such a way that there are exactly</em> <code>target</code> <em>neighborhoods</em>. If it is not possible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
<strong>Output:</strong> 9
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
<strong>Output:</strong> 11
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3
<strong>Output:</strong> -1
<strong>Explanation:</strong> Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == houses.length == cost.length</code></li>
	<li><code>n == cost[i].length</code></li>
	<li><code>1 &lt;= m &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= 20</code></li>
	<li><code>1 &lt;= target &lt;= m</code></li>
	<li><code>0 &lt;= houses[i] &lt;= n</code></li>
	<li><code>1 &lt;= cost[i][j] &lt;= 10<sup>4</sup></code></li>
</ul>
Array; Dynamic Programming
class Solution {
public:
    int minCost(vector<int>& houses, vector<vector<int>>& cost, int m, int n, int target) {
        int f[m][n + 1][target + 1];
        memset(f, 0x3f, sizeof(f));
        if (houses[0] == 0) {
            for (int j = 1; j <= n; ++j) {
                f[0][j][1] = cost[0][j - 1];
            }
        } else {
            f[0][houses[0]][1] = 0;
        }
        for (int i = 1; i < m; ++i) {
            if (houses[i] == 0) {
                for (int j = 1; j <= n; ++j) {
                    for (int k = 1; k <= min(target, i + 1); ++k) {
                        for (int j0 = 1; j0 <= n; ++j0) {
                            if (j == j0) {
                                f[i][j][k] = min(f[i][j][k], f[i - 1][j][k] + cost[i][j - 1]);
                            } else {
                                f[i][j][k] = min(f[i][j][k], f[i - 1][j0][k - 1] + cost[i][j - 1]);
                            }
                        }
                    }
                }
            } else {
                int j = houses[i];
                for (int k = 1; k <= min(target, i + 1); ++k) {
                    for (int j0 = 1; j0 <= n; ++j0) {
                        if (j == j0) {
                            f[i][j][k] = min(f[i][j][k], f[i - 1][j][k]);
                        } else {
                            f[i][j][k] = min(f[i][j][k], f[i - 1][j0][k - 1]);
                        }
                    }
                }
            }
        }
        int ans = 0x3f3f3f3f;
        for (int j = 1; j <= n; ++j) {
            ans = min(ans, f[m - 1][j][target]);
        }
        return ans == 0x3f3f3f3f ? -1 : ans;
    }
};