• [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;
        }
    };
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  • 原文地址:https://www.cnblogs.com/grandyang/p/5448505.html
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