> A: In this problem, the maximum length of the final sequence is 6, and we
> can easily find a basic method of depth-first-search to solve this problem 
> with a smaller scale. Since the enhancing levels can be divided into 11 
> categories: level 0~level 10, the first optimization is to select the best
> 6 skills in each category. By this optimization, the skill number N could
> be reduced to at most 60. The second optimization is also based on the 11
> categories of enhancing levels. We can find the rule that the first skill
> will always have no extra damage and the enhancing level of the last skill
> is useless. If the enhancing level of each skill in the sequence is 
> decided, it is easy to use greedy method to choose the best skills 
> available of each category to fit into the sequence. Since the last skill 
> has some special properties (we can choose it from any kind and greedy 
> method is wrong if we do not consider such properties), we can enumerate 
> the skill to fit into the last position, and enumerate the enhancing 
> level of the whole sequence, and then use greedy method to fill the blank 
> position with its best choice to figure out a solution. The best one 
> among these solution will be the final answer. 
> 
> B: 将这个简单多边形分成N个有向三角形（即面积可能是负数，就像在求简单多边形
> 的面积中一样），然后分别求每个三角形同圆公共面积。由于圆心在原点，所以可以
> 用原点作为划分三角形的“轴点”。
> 
> C: 搜索，每次尝试把一个三角形贴着当前未被覆盖区域的边界放置。很多剪枝可用，
> 例如如果当前边界上有一条边比剩余所有三角形中最短的边还短，那么就不需要继续
> 搜索
> 
> D: 我们用N = 2, W = {1，2}；M = 3, Q = (2, 4, 3) 来说明算法
> 首先建立3棵几乎相同状态树：
>         0                    0                     0
>        / \                  / \                   / \
>       1  (2)               1   2                 1   2
>      /   / \              /   / \               /   / \
>    (1)  1   2            1   1  (2)            1  (1)  2
>    /   /                /   /                 /   /  
>   1   1                1  (1)               (1)  1    
>  /                    /                     / 
> 1                   (1)                    1
> 区别在于每棵树都有一些节点被标记，表示这个节点对应的砝码堆的总重与Ri相同，
> 对于每个被标记的节点，向下一棵树的对应节点连一条权为0的边，于是M棵树连成
> 一个图
> 从第一棵树的0节点到最后一棵树的被标记节点的最短路长度即为所求
> 考虑到直接搜整个图（M棵树）所需要的队列可能存不下，所以可以一棵一棵树依次
> 进行搜索
> P.S：facer后来说这题可以进一步改编成加砝码和去砝码用的代价不同（例如取决于
> 砝码重量），模型相同，不过求最短路需要在树中进行DP
> 
> E: 每个formula都可以写成一个2-CNF，例如Xa AND Xb = 0 <=> ~Xa ∨ ~Xb，于是
> 题目可以转化为2-SAT问题
> 
> F: 先搜出来长度不超过T的所有可能路线。然后枚举Tom的起点，计算和并且求出最
> 大值
> 
> G: 最小费用最大流。 离散化每个ai,bi，排序后,变为v1,v2,v3...,vg
> 增加源点S和汇点T， S到v1增加容量为K,费用为0的边
> vg到T增加容量为K,费用为0的边， vi到vi+1增加一条容量为K,费用为0的边
> 对于题目中给定的线段， 增加ai到bi容量为1,费用为-wi的边
> 负的最小费用最大流求出的最小费用即为所求
> P.S： 比赛中的时限比较紧，只有标程的2倍多；比赛之后放宽了
> 
> H: 枚举两条平行y轴的直线, 对两条直线中的点用双端队列处理, 找出最小值
> 时间复杂度 O(N^3)