久久久噜噜噜久久熟女AA片,久久丝袜精品中文字幕,亚洲人成电影网站久久

[Retrospect] Andrew Stankevich's Contest

FreePeter — Thu, 01 May 2008 12:45:00 GMT

Retrospect only, some algorithms may be wrong...
Warning : 剧透慎�?..

Andrew Stankevich's Contest #1
Chinese Girls' Amusement
推结论直接算
Reactor Cooling
有上下界的环形流
New Year Bonus Grant
�l�典的树形DP
Matrix Multiplication
化简一下推�l�论
Nice Patterns Strike Back (Recommend)
状态压�~�DP后用matrix来优化，或者用上次monthly时LK的方法�?br id="cyo319"> Get Out! (Recommend)
可以用那个生成环的基的DFS,�c�M��japan那道题�?br id="cyo322"> Beautiful People
最长虾�c�_��列，8�q�稍微有个细节要�?br id="cyo325"> Cracking' RSA (Recommend)
求bool方程�l�的解数。。。也��是求下自由变量个数

Andrew Stankevich's Contest #2
Non Absorbing DFA
记得是个很正常的DP�?br id="cyo335"> The Towers of Hanoi Revisited
�l�典的n个塔的hanoi�Q�记得要DP...
Hyperhuffman
��是huffman问题吧，已经排好序，可以O(n)的�?br id="cyo341"> Little Jumper (Recommend)
不错的物理题
首先可以惌��成两只青蛙一起从两边跟�?br id="cyo345"> 主要问题��是计算�l�定一个v后的青蛙可达区间�?br id="cyo346"> 取到最值只�?�U�情况：从上面擦�q�，从下面擦�q�，45度�v跻I��如果可以�Q?br id="cyo347"> Quantization Problem
又是一个正常的DP...
Roads
�l�典问题了�?br id="cyo353"> 生成树上的权值必定是减少�Q�其他边上的权值必定增大，讑օ�分别为ai, bj
然后对�Q意一条不在生成树上的边，加入到生成树上�Ş成一个环�Q�然后这条边的权值应�?gt;=环上所有边的权倹{�?br id="cyo355"> 然后我们可以列出一堆�Ş如ai + bj >= 一个正数的不等式。。。然后。。。km��法的标��~~~
详情可以看km��法的证明~~~
Robbers
首先令k[i] = m * x[i] / y�Q�取下整�Q�然后可能k[i]的和不到m�Q�要增加一些k[i]�Q�当然是�Q�每�ơ找增大后delta最��的~~~
主要是，��g��可以用heap优化为O(nlogn)。。�?br id="cyo361"> Toral Tickets (Recommended)
比较��奇的Polya

Andrew Stankevich's Contest #3
Areas (Recommended)
我的做法是基于半�q�面交的�Q�对每条直线�Q�枚举��用它左边的半�q�面�q�是双��的半�q�面�Q�最后如果是一个有限��^面则�q�回。。�?br id="cyo370">注意搜烦�q�程中当半��^面被切空后就可以return�Q�由于最后只有O(n^2)块，所以复杂度是O(n^4)的（使用O(n^2)的半�q�面交）
SGU上时限很宽，ZJU上这么做旉��有点紧（�?.95s AC�?_-bbbbbbb�Q?br id="cyo372"> Beloved Sons
按偏��q��度从大到��找增广路跑max_match��可以了。。�?br id="cyo375"> Strange Counter
构�?br id="cyo378"> �l�护�q�么一个性质两个2之间臛_��有一�?。。。然后。。。讨论。。�?br id="cyo379"> Data Transmission (In List)
据说是预��?+ 使劲优化...(by Lunarmony)
Strong Defence
嗯，首先颜色��C��会超�q��Q何一条�\的长度是把。。。所以颜色数臛_��是最短�\的长度�?br id="cyo385"> 然后我们跑dijstra的时候把边着上颜色就是了。。�?br id="cyo386"> Weird Dissimilarity
�l�典的DP
PL/Cool
据说是模�?from oibh)。。�?br id="cyo392"> Royal Federation (In List)
据说是构�? not AC yet.
Two Cylinders
写出�U�分式后romberg.

Andrew Stankevich's Contest #4
The Smart Bomb
��单的推一下�?br id="cyo3104"> I Just Called ...
模拟�Q�要用Trie树�?br id="cyo3107"> Order-Preserving Codes
模仿huffman那样�Q�只是每�ơmerge盔R��的�?br id="cyo3110"> More Divisors
�l�典的DP, f[i][j]用前i个素数得到j个约数的最��数。。�?br id="cyo3113"> Long Dominoes
状态压�~�DP
The Magic Wheel
应该选择�W�一个点�Q�然后寻找下一层的两个方向最�q�的都试一下就行了。O(N)
Cracking SSH
DP...
Periodic Tilings
好像某年final有类似的题。应该有�l�论的说�?br id="cyo3125"> Not AC yet
Trade (In List)
Not AC yet, 可以看看
Counting Triangulations (Recommended)
一道还��不错的DP题，8�q�题目描�q�好像有点不清我记得�?br id="cyo3132"> Unfair Contest
搜烦+模拟

Andrew Stankevich's Contest #5
Unique Attack (Recommended)
判断最��割是否唯一的题�Q�就是用两种�Ҏ(gu��)��构造是否一栗��?br id="cyo3142"> Burning Bridges
ms是很�l�典的用桥来作的�?br id="cyo3145"> Circles
�l�典的��^面图�Ƨ拉公式�?br id="cyo3148"> Linear Programming Dual
好象是线性规划，Not AC yet
DVD (In List)
相当�Ҏ(gu��)��写错的DP题�?br id="cyo3154"> Think Positive
记得可以O(n)扫描的�?br id="cyo3157"> Ranking
�ȝ��的模拟题�?br id="cyo3160"> Driving Straight
也是很经典的思�\了，先DP(或曰BFS)。然后走一遍，在满��x��解的前提下尽量往那个方向走�?br id="cyo3163">
Andrew Stankevich's Contest #6
Ackerman's Function (Recommended)
可以认�ؓ是找规律
The Minimal Angle
记得要O(n)�Q�取�q�_��数还是什么都可以�?br id="cyo3172"> Yellow Code
我记得还是比较容易YY一个构造的。。�?br id="cyo3175"> Yet Another Digit
DP吧�?br id="cyo3178"> Graduated Lexicographical Ordering (In List)
�c�M��于vietnam的那道题。。。相当麻烦。。。徏议实��C��。。�?br id="cyo3181"> GSM
高精度开斚w��。或者打表～
Warehouse Keeper (In List)
KM,8�q�我记得�Ҏ(gu��)��T?
Don't Go Left
记得又是一个状态机的BFS�?br id="cyo3190"> Railroad Sort
很有意思的构造，大体思�\是每�l�过一个station�Q�留住后一半，放行前一半。。。n个正好给2^n个数排序�?br id="cyo3193">
Andrew Stankevich's Contest #7
Little Brackets
�l�典的dp, NOI陨石的秘密简化版。。�?br id="cyo3199"> f[n][k] = n�Ҏ(gu��)��P��<=k�?br id="cyo3200"> f[n][k] = sigma(f[m][k] * f[n - m - 1][k - 1])�Q�输出f[n][k] - f[n][k - 1]
��是每次��d��一整个括号块�?br id="cyo3202"> Under Control (In List)
转换坐标��L��化吧
�c�M��思�\有道��复杂�?a id="cyo3206">Soldier
Holidays (In List)
Not AC Yet
Laboratory
记得列出式子调整下就行了。时限吓人的�?br id="cyo3213"> Maps
Crazy Painter
Puzzle
没记错的话BFS一下吧。。�?br id="cyo3220"> Quest
�l�典的状态压�~�BFS�Q�输�Ҏ(gu��)��有点�ȝ��。。�?br id="cyo3223"> Stable Sets

FreePeter 2008-05-01 20:45 发表评论

[Solution] Japan 2007

FreePeter — Thu, 01 May 2008 12:25:00 GMT

[Solution] Tokyo 2007

And Then There Was One
�l�典题，递推�?br id="vzoi3">
Prime Gap
��单题

Minimal Backgammon
DP

Lowest Pyramid
比较�ȝ��的题目，大体做法是枚举一个点�Q�根据距��ȝ��{�可以枚丑֏�一个点�Q�这些点很少了）�Q�然后确定下最后一个点�Q�check.

Geometric Map
比较�ȝ��的预处理 + dijstra

Slim Span
�l�典问题了，按边大小排序�Q�每�ơ加一条边�Q�如果�Ş成环��L��环上最��的辏V��check

The Morning after Halloween
BFS�Q�最后用A*�q�掉的。用max(当前位置到目标位�|?��C�h(hu��n)

Bug Hunt
��单模�?br id="vzoi24">
Most Distant Point from the Sea
可以用二分＋半��^面交�?br id="vzoi27">但也可以惌��所有边朝里面挤压，�q�样最后要么是两条边压��C��P��要么三条边压成一个点。O(n^3)枚�D�?br id="vzoi28">
The Teacher's Side of Math
注意到p,q都是质数�Q�所以答案是0必须是所有其他项�p�L��全�ؓ0�Q�这样就可以解方�E�了�?br id="vzoi31">用long double + 最大主元法可以�q��?br id="vzoi32">

FreePeter 2008-05-01 20:25 发表评论

[Solution] SWERC 2007 Southwestern Europe

FreePeter — Thu, 01 May 2008 12:24:00 GMT

[Solution] SWERC 2007 Southwestern Europe
比较��单，有些题读题比较郁闗��?br id="s_1y1">
BEATBIT
两DFA是否同构�Q�bfs or 判断树是否同构都可以�Q�因��Z��证了可以�l�止所以没有环存在�Q?br id="s_1y5">
Prester John
题意没说清，走�\的方式类��g��NFA, BFS��p��了�?br id="s_1y9">不过可以��Z��数据让所有程序T...

Robotruck
O(N*C)的DP

Jumping Hero
BFS�Q�最�?00 * 300 * 5000 * 5�U�状态，当然实际上远�q�不到�?br id="s_1y18">
Board Game
Bellman-Ford

The Bridges of Kolsberg
�l�典DP

The Finest Chef
最优权匚w��

IP-TV
MST

Ladies' Choice
�E�_��婚姻

FreePeter 2008-05-01 20:24 发表评论

[Solution] Dhaka 2007

FreePeter — Thu, 01 May 2008 12:22:00 GMT

[Solution] Dhaka 2007

Bachelor Arithmetic
�U�杀�?/span>

Nested Squares
模拟�?/span>

The Dumb Grocer
首先要有1是吧。。。然后我们按�?/span>1的个数来分类�Q�我们来计算恰有k�?/span>1的方案数�?/span>
我们�?/span>k�?/span>1的基��上加入新的数�Q�显然第一个数只能�?/span>k+1
然后加入的数只能�?/span>k + 1 or 2 * (k + 1)
如法炮制。。。发现非1的数都具�?/span>(k + 1) * t的�Ş式。。。设其依�ơ�ؓ(k + 1) * ti
�?/span>{ti}�q�些��C��满��题目的性质。。。共�?/span>f((n - k) / (k + 1))�U�方案�?/span>

�?/span>f(n)是要求的函数�Q�则f(n) = sigma(f((n - k) / (k + 1)), (k + 1) | (n + 1) , k>=1
f(0) = 1
�q�样直接做会T...
我们�?/span>g(n) = f(n - 1)
�?/span>g(n) = f(n - 1) = sigma(f((n - k - 1) / (k + 1))) = sigma(f(n / (k + 1) - 1)) = sigma(g(n / (k + 1)), (k + 1) | n, k >= 1
�?/span>n = p1^a1 * p2^a2 * ... * pr*ar
�?/span>h(p1, p2,.., pr, a1, a2...ar) = g(n)
= h(p1,p2, ...pr, b1, b2, ...br),
0<=bi <= ai, bi不全=ai
注意对于一个确定的n�Q?/span>h()中的p1, p2...pr在计��过�E�中始终不变。。。所以。。�?strong>计算�l�果�?/strong>pi无关,只与ai有关
�q�样状态数��大大减��了。。。直接因式分解后dp��p��了。。�?/span>

ACM Puzzles
状态压�~?/span>dp

�?/span>The Bells are Ringing�?/span> Photographic Tour
�q�三题好像当时没�?/span>summary。。。所以我们假设比较简单～～～

You are around me ...
首先旋�{坐标�Q�变成��^行与xy轴的椭圆�Q�然后坐标�׾~�。。。变成圆。。。最�q�点寏V��。。脓(chu��ng)模板。。�?/span>
ZJU2107 Quoit Design 一道测最�q�点对的题�?/span>

Infinite Matrix
昄��Q�对于固定的j, Ri, j是一个关�?/span>i的多��式�?/span>
注意到数�?/span>Ri, j的差分序�?/span>R(i + 1, j) - R(i, j)是可以求出来的（利用Mj, k <= 10的条�Ӟ��可以�?/span>O(10*n^2)的时间内��出�Q?/span>
然后有了差分序列求通项��是O(n^2)的事情�?/span>
然后�?/span>S(p, j)(n) = Sigma(i^p * R(i, j)) i <= n
�l�箋利用差分序列之类的方法求�q�个�Q�最后再求一�?/span>Sum_S(p, j)
预处理复杂度大致�?/span>O(10*n^3)的�?/span>
后面的就好办了，�Ҏ(gu��)��个询问，�?/span>(i + 1)^p二项式展开�Q�最�?/span>10��，然后利用公式直接计算�?/span>
处理询问复杂�?/span>O(p*q*n)
POJ3529 Matrix Analysis�Q�类似的思想�Q�更��单～

Magnetic Train Tracks
�l�定n个点�Q�求可以构成多少个锐角三角�Ş�?/span>
n <= 1200
话说求锐角三角�Ş不太好算是吧。。。补集�{换，我们来求钝角/直角三角�?/span> <=> 求钝�?/span>/直角个数。。�?/span>
后面的事情就��单了�Q�是�Ҏ(gu��)��个点�Q�将其他�Ҏ(gu��)��照极角排�?/span> + 扫描�?/span>
Dhaka 2005 Counting Triangles 也是一道补集�{换的题～�Q��{化成求三点共�U�的个数�Q?/span>
Shanghai 2004 Amphiphilic Carbon Molecules 也是一道极角排�?/span>+扫描的题�?/span>

FreePeter 2008-05-01 20:22 发表评论

TCO Round1, [500], CHOMP Game...

FreePeter — Sat, 16 Feb 2008 20:15:00 GMT

500分的题。。�?br>�׃��之前看到�q�chomp game�Q?《Game Theory》的�l�习里有)�Q�然后开始试图推公式之类的。。。在wiki上找到rectangle情况先手必胜的证明：

Who wins?

Chomp belongs to the category of impartial 2-player perfect information games.

It turns out that for any rectangular starting position bigger than 1 × 1 the 1st player can win. This can be shown using a strategy-stealing argument: assume that the 2nd player has a winning strategy against any initial 1st player move. Suppose then, that the 1st player takes only the bottom right hand square. By our assumption, the 2nd player has a response to this which will force victory. But if such a winning response exists, the 1st player could have played it as his first move and thus forced victory. The 2nd player therefore cannot have a winning strategy.

Computers can easily calculate winning moves for this game on two-dimensional boards of reasonable size.

很优��的证明。。。只可惜不能提供��M��strategy...-_-bbbbbbbb
最后终于�?zh��n)�出来�q�题规定��盘3*n, n<=100,所以就100*100*100的dp��p��了。�?_-bbbbbbbbbb

p.s. wiki : Chomp Game

p.s. ��实觉得一知半解是一个很�Ҏ(gu��)��出错的情�?..因�ؓ如果完全不知道思维也就没有��M��限制�?曄��看到�q�么...感觉会有点紧�?惌��赶紧搞掉的��U�感�? & 试图用记忆中的套路去�?..但有时候可能没有关�p?例如�q�个game, 先手必胜的证明�ƈ不能提供��M��先手如何operate的信�?..,如果�l�箋往�q�个上面惛_��直接挂了...-_-bbbbbbb)
�q�有��是有可能会出现�c�M��?当时��Z��么不仔细推清�?之类的念�?..�q�个seems�Ҏ(gu��)��解决...

感觉如果是完全陌生的题想法通常�Ҏ(gu��)��比较open, 如果感觉�q�个模型熟�?zh��n)�一般都会试囑־�熟�?zh��n)�的模型上�?..大多数情况下�q�样��实可以节省旉��...但是如果失去了open的思维 + 熟�?zh��n)�的模型无法解军_��orz�?..

FreePeter 2008-02-17 04:15 发表评论