256. Paint House

最新推荐文章于 2020-07-27 21:40:21 发布

原创最新推荐文章于 2020-07-27 21:40:21 发布 · 264 阅读

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#leetcode #dynamic programming

本文详细解析了PaintHouse问题的动态规划解决方案，通过分析不同颜色涂色的成本，找到使所有房屋颜色不同且总成本最低的策略。介绍了动态规划的基本概念，展示了如何利用前一状态的最小成本来决定当前状态的成本，最终求得全局最优解。

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256. Paint House

- 易错点
- code

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.

解题思路：

一道很明显的动态规划的题目. 每个房子有三种染色方案, 那么如果当前房子染红色的话, 最小代价将是上一个房子的绿色和蓝色的最小代价+当前房子染红色的代价. 对另外两种颜色也是如此. 因此动态转移方程为:

Sub-problem: find the minimum cost to paint the houses up to current house in red, blue or green.
Function:
Red: min(f[i - 11][1], f[i - 1][2]) + costs[i][0].
Blue: min(f[i - 1][0], f[i - 1][2]) + costs[i][1].
Green: min(f[i - 1][0], f[i - 1][1]) + costs[i][2].
Initialization: f[0][i] = 0.
Answer: min(f[costs.length][i]).

Line 933: Char 34: runtime error: reference binding to misaligned address 0xbebebebebebebebe for type 'value_type', which requires 4 byte alignment (stl_vector.h)
0xbebebebebebebebe: note: pointer points here
<memory cannot be printed>

这个错误出现是因为costs[i]超出了边界

易错点

dp比costs多出一维，应该使用 costs[i - 1] 否则会报上面的错误

code

class Solution {
public:
    int minCost(vector<vector<int>>& costs) {
        int n = costs.size();
        // 直接initialize成0更好
        vector<vector<int>> dp(n + 1, vector<int>(3, INT_MAX));
        
        for (int i = 0; i < 3; i ++)
            dp[0][i] = 0;
        
        for(int i = 1; i < n + 1; i++){
            dp[i][0] = min(dp[i - 1][1], dp[i - 1][2]) + costs[i - 1][0];
            dp[i][1] = min(dp[i - 1][0], dp[i - 1][2]) + costs[i - 1][1];
            dp[i][2] = min(dp[i - 1][0], dp[i - 1][1]) + costs[i - 1][2];
        }
        
        return min(dp[n][0], min(dp[n][1], dp[n][2]));
    }
};