lintcode minimum-adjustment-cost 最小调整代价

问题描述

lintcode

笔记

假设buff[i][j]的含义为：将第i个数改成j的调整代价。如果改成j之后不符合调整的目标则buff[i][j]=INT_MAX。

初始化缓存将第0个数改为j的代价是

buff[0][j] = abs(A[0]-j);

为了符合调整目标，将第i个数改成j的话，第i-1个数k的可取值范围是[j-target, j+target]。在符合调整目标的情况下的状态转移方程为：

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

代码

class Solution {
public:
/**
* @param A: An integer array.
* @param target: An integer.
*/
int MinAdjustmentCost(vector<int> A, int target) {
// write your code here
const int len = A.size();
int buff[len][101];
for (int j = 1; j <= 100; j++)
{
buff[0][j] = abs(j-A[0]);
}
for (int i = 1; i < len; i++)
{
for (int j = 1; j <= 100; j++)
{
buff[i][j] = INT_MAX;
int lo = max(1, j - target);
int hi = min(100, j + target);
for (int k = lo; k <= hi; k++)
{
buff[i][j] = min(buff[i][j], buff[i-1][k]+abs(j-A[i]));
}
}
}
int res = INT_MAX;
for (int i = 1; i <= 100; i++)
{
res = min(res, buff[len-1][i]);
}
return res;
}
};