2017 ACM-ICPC World Finals
ABCDEFGHIJKL
复盘：fz
开场看到A是个几何题,就直接交给叶队了.
看了一发F,是个签到题,直接就写了.
0:13 F(+0)
期间阿爷会了E和I两个签到题,写完都过了.
0:16 E(+0)
0:19 I(+0)
叶队说会了A，但是细节很复杂.写了一会之后我猜了一个L的结论给R爷.然后R爷开始写L.
R爷写L的时候我看了一会D和K，感觉都不会.K大概猜和Border字典序有关,但是不太敢确定.然后会了C.
R爷写完L之后wa 21了,改了几个错误之后还是wa 21,感觉算法没错,就让R爷先写C，没考虑到一列都是0的情况.
01:53 C(+1)
然后叶队开始继续写A,和R爷眼看了一会L之后R爷发现自己有个地方写错,写完过了.
02:16 L(+2)
然后和R爷讨论了一会D和K，D没什么想法,K需要打表.大概过了40分钟之后打了个表,发现应该是字典序相关,写了一发过了.
03:26 K(+0)
最后选择开了G，然后拿着D,G一起想.叶队继续调A。
把D画成两个单调序列之后就觉得可能有单调性.推了一发发现果然有,交给R爷写了一发.
04:37 D(+0)
最后帮叶队看A，换了若干发姿势都没过,GG.
A：
B：处理出每个点必定选,一定不选,以及三张里面至少有一张的集合.算出每个人的可能集合.枚举三张牌直接暴力一下.
C：考虑要最大化能够共用的行列数量,转化为二分图最大权匹配.
D：考虑最优解一定在A的单调下降和B的单调下降序列上.那么推测其有单调性,推一发即可.
E：二分。
F：预处理转移函数dp.
G：由于每轮扩展都会使得其行列-2,考虑一个逆着求原来形状的过程,那么我们只需要令new[i][j]=last[i-1][j-1]^...另外八个格子的new值,就可以得到一张新图.同时,如果某个位置被xor上了1，那么事实上它会对第i+1行的求解产生影响,并且肯定是两个错夹杂一个对的.于是我们行和列各做一遍求出其位置,将其删去,再做一遍如果还有就不行,否则就更新下去.
H：
I：模拟。
J：
K：打表猜结论得与其[n-m+1,m]这个后缀中的border字典序有关.
L：从x高往低确定每个点,每个点一定是选择y最小的最优.匹配出方案之后判断是否合法.
总结：
(by fz)：
整体节奏问题不是特别大.D,K过的晚确实是因为需要这么多时间去想或者打表.
中期40分钟空闲的时候应该把B,H,J都开出来比较好,至少把所有不超过一页的题都读一遍.B其实并不难，但是被榜带的比较歪.
summary by yjn:
I should have a geometry template......
Never use inverse trigonometric function if you do not neccesarily need to know what the angle exactly is.
Specially, do not take atan2 to do any comparisive work.
The precision problem really matters. Use Interger operations if you can.
When you need to determine if two segments intersects(not strictly), check their boxes.
Dealing issues about segment and non-convex polygon is tricky, now I'm much more experienced.
Again, I should have a template. I'm working on it.
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