[LintCode] Minimum Adjustment Cost

Minimum Adjustment Cost

Given an integer array, adjust each integers so that the difference of every adjacent integers are not greater than a given number target.

If the array before adjustment is A, the array after adjustment isB, you should minimize the sum of |A[i]-B[i]|

Example

Given [1,4,2,3] and target = 1, one of the solutions is [2,3,2,3], the adjustment cost is 2 and it's minimal.

Return 2.

Note

You can assume each number in the array is a positive integer and not greater than 100.

动态规划，dp[i][j]表示把第i个数调整为j后的最小代价，状态转移方程为：

dp[i][j] = min(dp[i][j], dp[i-1][k] + abs(A[i] - j));

其中，k表示上一个数调整后的数值，范围在max(0, j - target) 与 min(100, j + target) 之间。

 1 class Solution {

 2 public:

 3     /**

 4      * @param A: An integer array.

 5      * @param target: An integer.

 6      */

 7     int MinAdjustmentCost(vector<int> A, int target) {

 8         // write your code here

 9         if (A.empty()) return 0;

10         vector<vector<int>> dp(A.size(), vector<int>(101, INT_MAX));

11         for (int i = 0; i <= 100; ++i) {

12             dp[0][i] = abs(A[0] - i);

13         }

14         for (int i = 1; i < A.size(); ++i) {

15             for (int j = 0; j <= 100; ++j) {

16                 for (int k = max(0, j - target); k <= min(100, j + target); ++k)

17                     dp[i][j] = min(dp[i][j], dp[i-1][k] + abs(A[i] - j));

18             }

19         }

20         int res = INT_MAX;

21         for (int i = 0; i <= 100; ++i) {

22             res = min(res, dp[A.size()-1][i]);

23         }

24         return res;

25     }

26 };