# 256. Paint House 粉刷房子

@TOC

## # 题目描述

There are a row of n houses, each house can be painted with one of the three colors: `red, blue or green`. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.

The cost of painting each house with a certain color is represented by a `n x 3` cost matrix. For example, `costs[0][0]` is the cost of painting house 0 with color `red`; `costs[1][2]` is the cost of painting house 1 with color `green`, and so on... Find the minimum cost to paint all houses.

Note: All costs are positive integers.

Example:

``````Input: [[17,2,17],[16,16,5],[14,3,19]]
Output: 10
Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue.
Minimum cost: 2 + 5 + 3 = 10.
``````

## # 解题方法

### # 动态规划

``````dp[n][0] = min(dp[n - 1][1], dp[n - 1][2]) + costs[n - 1][0];
dp[n][1] = min(dp[n - 1][0], dp[n - 1][2]) + costs[n - 1][1];
dp[n][2] = min(dp[n - 1][0], dp[n - 1][1]) + costs[n - 1][2];
``````

C++代码如下：

``````class Solution {
public:
int minCost(vector<vector<int>>& costs) {
const int N = costs.size();
vector<vector<int>> dp(N + 1, vector<int>(3, 0));
int res = 0;
for (int n = 1; n <= N; ++n) {
dp[n][0] = min(dp[n - 1][1], dp[n - 1][2]) + costs[n - 1][0];
dp[n][1] = min(dp[n - 1][0], dp[n - 1][2]) + costs[n - 1][1];
dp[n][2] = min(dp[n - 1][0], dp[n - 1][1]) + costs[n - 1][2];
}
return min(dp[N][0], min(dp[N][1], dp[N][2]));
}
};
``````

## # 日期

2019 年 9 月 18 日 —— 今日又是九一八