256. Paint House 粉刷房子
- 作者: 负雪明烛
- id: fuxuemingzhu
- 个人博客:http://fuxuemingzhu.cn/
@TOC
题目地址:https://leetcode-cn.com/problems/paint-house/
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.
题目大意
假如有一排房子,共 n 个,每个房子可以被粉刷成红色、蓝色或者绿色这三种颜色中的一种,你需要粉刷所有的房子并且使其相邻的两个房子颜色不能相同。
当然,因为市场上不同颜色油漆的价格不同,所以房子粉刷成不同颜色的花费成本也是不同的。每个房子粉刷成不同颜色的花费是以一个n x 3的矩阵来表示的。
例如,costs[0][0] 表示第 0 号房子粉刷成红色的成本花费;costs[1][2]表示第 1 号房子粉刷成绿色的花费,以此类推。请你计算出粉刷完所有房子最少的花费成本。
解题方法
动态规划
计算刷完第n房子为止,三种颜色分别累计需要的最少花费dp[n][0], dp[n][1], d[n][2]。那么递推公式是:
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];
意义是不能和前面的房子刷相同的颜色,那么就使用前面房子另外两种颜色的累计最小花费,加上刷当前的颜色的花费。最后去最小即可。
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