poj 2559 DP - 推酷



我想到的是最大的矩形，中间一定有个最矮的某个单位矩形，所以用两个数组记录histogram[i]左右两边第一个比它小的单位矩形的序号leftLowerId[i]和rightLowerId[i]，那么对于histogram[i]，它自己的最大矩形面积就是(rightLowerId[i] - leftLowerId[i] - 1) *  histogram[i]。

  这里找leftLowerId和rightLowerId的时候用DP加速。以rightLowerId为例，找到右边比histogram[i]矮的矩形，停止，遇到比histogram[i]高的矩形j，直接跳到比histogram[j]矮的矩形rightLowerId[j].


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