Paint House

想想想

public class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0) {
    		return 0;
    	}  
    	for (int i = 1; i < costs.length; i++) {
    		costs[i][0] = costs[i][0] + Math.min(costs[i - 1][1], costs[i - 1][2]);
    		costs[i][1] = costs[i][1] + Math.min(costs[i - 1][0], costs[i - 1][2]);
    		costs[i][2] = costs[i][2] + Math.min(costs[i - 1][0], costs[i - 1][1]);
    	}
    	return Math.min(costs[costs.length - 1][0], Math.min(costs[costs.length - 1][1], costs[costs.length - 1][2]));
    }
}


B. Array Recoloring time limit per test2 seconds memory limit per test256 megabytes You are given an integer array a of size n . Initially, all elements of the array are colored red. You have to choose exactly k elements of the array and paint them blue. Then, while there is at least one red element, you have to select any red element with a blue neighbor and make it blue. The cost of painting the array is defined as the sum of the first k chosen elements and the last painted element. Your task is to calculate the maximum possible cost of painting for the given array. Input The first line contains a single integer t (1≤t≤103 ) — the number of test cases. The first line of each test case contains two integers n and k (2≤n≤5000 ; 1≤k<n ). The second line contains n integers a1,a2,…,an (1≤ai≤109 ). Additional constraint on the input: the sum of n over all test cases doesn't exceed 5000 . Output For each test case, print a single integer — the maximum possible cost of painting for the given array. Example InputCopy 3 3 1 1 2 3 5 2 4 2 3 1 3 4 3 2 2 2 2 OutputCopy 5 10 8 Note In the first example, you can initially color the 2 -nd element, and then color the elements in the order 1,3 . Then the cost of painting is equal to 2+3=5 . In the second example, you can initially color the elements 1 and 5 , and then color the elements in the order 2,4,3 . Then the cost of painting is equal to 4+3+3=10 . In the third example, you can initially color the elements 2,3,4 , and then color the 1 -st element. Then the cost of painting is equal to 2+2+2+2=8 . 用cpp解决
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03-19
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