paint i 和 ii一樣，只是限定了k=3，所以可以寫的簡單一些。

paint ii
https://leetcode.com/problems/paint-house-ii/
未壓縮空間：

class Solution {
public:
    int minCostII(vector<vector<int>>& costs) {
        int n = costs.size();
        if(n==0) return 0;
        int m = costs[0].size();
        vector<vector<int> > dp(n,vector<int>(m,0));
        //init
        for(int i = 0; i < m; i++)
            dp[0][i] = costs[0][i];
        
        //dp
        for(int i = 1; i < n; i++) {
            for(int j = 0; j < m; j++) {
                dp[i][j] = INT_MAX;
                for(int k = 0; k < m; k++) {
                    if(j == k) continue;
                    dp[i][j] = min(dp[i][j], dp[i-1][k] + costs[i][j]);
                }
            }
        }
        
        int res = INT_MAX;
        for(int i = 0; i < m; i++)
            res = min(res, dp[n-1][i]);
        return res;
    }
};

壓縮空間是否可行？不能直接壓縮，要記錄兩個min1和min2：
https://leetcode.com/discuss/52982/c-dp-time-o-nk-space-o-k

class Solution {
public:
    int minCostII(vector<vector<int>>& costs) {
        if (costs.empty()) return 0;
        int n = costs.size(), k = costs[0].size(), m1 = 0, m2 = 0;
        vector<int> dp(k, 0);
        for (int i = 0; i < n; i++) {
            int t1 = m1, t2 = m2;
            m1 = m2 = INT_MAX;
            for (int j = 0; j < k; j++) {
                dp[j] = (dp[j] != t1 ? t1 : t2) + costs[i][j];
                if (m1 <= dp[j]) m2 = min(m2, dp[j]);
                else m2 = m1, m1 = dp[j];
            }
        }
        return m1;
    }
};

Paint Fence
https://leetcode.com/problems/paint-fence/
不能有兩個以上相鄰的fence是相同color。

第三根柱子要么根第一個柱子不是一個顏色，要么跟第二根柱子不是一個顏色

class Solution {
public:
    int numWays(int n, int k) {
        if(n==0 ||k==0) return 0;
        vector<int> dp(n);
        dp[0] = k;
        dp[1] = k*k;
        for(int i = 2; i < n; i++) {
            dp[i] = (k-1)* (dp[i-1] + dp[i-2]);
        }
        return dp[n-1];
    }
};