seam

难度飙升。需要非常注意优化问题，真的让最短路径算法直接刻进大脑了：

除了官方的FAQ提示，我这里列一下个人经验：

最好不要存picture成员变量，因为会使用多个getRGB()方法（尽管getRGB()方法比get()更快，但是多次调用会在Timing这里丢分）个人使用int[][] color来存储RGB值
RGB值占4 Bytes，各字节分别是alpha, red, green, blue
行优先访问比列优先访问要快，为什么？详见Cache（据说CSAPP的LAB有提及）
多熟悉最短路径算法：Dijkstra/Bellman-Ford/Topological Order 这个实验用到后者

Official Tips:

Don’t use an explicit EdgeWeightedDigraph. Instead, execute the topological sort algorithm directly on the pixels.
When finding a seam, call energy() at most once per pixel. For example, you can save the energies in a local variable energy[][] and access the information directly from the 2D array (instead of recomputing from scratch).
Avoid redundant calls to the get() method in Picture. For example, to access the red, green, and blue components of a pixel, call get() only once (and not three times).
The order in which you traverse the pixels (row-major order vs. column-major order) can make a big difference.
Creating Color objects can be a bottleneck. Each call to the get() method in Picture creates a new Color object. You can avoid this overhead by using the getRGB() method in Picture, which returns the color, encoded as a 32-bit int. The companion setRGB() method sets the color of a given pixel using a 32-bit int to encode the color.
Don’t explicitly transpose the Picture or int[][] until you need to do so. For example, if you perform a sequence of 50 consecutive horizontal seam removals, you should need only two transposes (not 100).
Consider using System.arraycopy() to shift elements within an array.
Reuse the energy array and shift array elements to plug the holes left from the seam that was just removed. You will need to recalculate the energies for the pixels along the seam that was just removed, but no other energies will change.

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