[LintCode] Paint House II 粉刷房子之二

There are a row of n houses, each house can be painted with one of the k colors. 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 k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

Notice

All costs are positive integers.

Example
Given n = 3, k = 3, costs = [[14,2,11],[11,14,5],[14,3,10]] return 10

house 0 is color 2, house 1 is color 3, house 2 is color 2, 2 + 5 + 3 = 10

LeetCode上的原题，请参见我之前的博客Paint House II。

class Solution {
public:
    /**
     * @param costs n x k cost matrix
     * @return an integer, the minimum cost to paint all houses
     */
    int minCostII(vector<vector<int>>& costs) {
        if (costs.empty() || costs[0].empty()) return 0;
        int m = costs.size(), n = costs[0].size();
        int min1 = 0, min2 = 0, idx1 = -1;
        for (int i = 0; i < m; ++i) {
            int m1 = INT_MAX, m2 = m1, id1 = -1;
            for (int j = 0; j < n; ++j) {
                int cost = costs[i][j] + (j == idx1 ? min2 : min1);
                if (cost < m1) {
                    m2 = m1; m1 = cost; id1 = j;
                } else if (cost < m2) {
                    m2 = cost;
                }
            }
            min1 = m1; idx1 = id1; min2 = m2;
        }
        return min1;
    }
};

本文转自博客园Grandyang的博客，原文链接：粉刷房子之二[LintCode] Paint House II ，如需转载请自行联系原博主。