Those rules are describing the source file that you submit. In particular you can name your solution things like "my solution.cpp", "naïve.cpp", "n^2.cpp".

В первой задаче зашли хэши. Добавляем буквы в ответ по одной, считаем хэши префиксов. Если длина ответа больше, чем длина вырезаемой строки, то хэшами проверяем совпадение. Чтобы не делать операции удаления из строки-ответа, я записывал ответ в массив char'ов и при удалении строки передвигал указатель на текущий символ назад.

В третьей задаче можно заметить, что при соединении игравших друг с другом команд получится дерево. Нам нужно найти остовное дерево максимального веса, для чего достаточно модифицировать сортировку в алгоритме Крускалла.

А 1 в серебре, такая же как и в бронзе (есть текст и строка, надо удалять вхождения строки в текст, учитывая появление новых вхождений и |s|<=10^6 |t|<=100)?

Yes we can discuss the problems now

Well, I believe USACO's main goal is selection and training for IOI and IOI has full feedback, so their decision was quite logical.

Usually it takes a day or two :)

For the second one you need an algorithm for multiple pattern matching (I used Aho Corasick)

For censor, you can precompute the hashes for each substring of the N censored words. Then when doing the deletion, you can maintain pointers so that you can recompute hashes. I didn't do this during the contest, I did a KMP which received 12/15

Third problem can be solved using sqrt decomposition:

1) we should group all requests to about sqrt(n) groups,

2) before processing each group we should build convex hull with O(n) time (all points we can sort before processing requests),

3) before processing each group we should sort all lines by angle and process them with convex hull with pointers,

4) and finally for each line we should process new points from block of requests.

So solution has O(n sqrt(n) + n log(n)) complexity.

Link to problem(s)?

For problem 3, consider the complete graph where the nodes are the teams and the edges are the potential scores if they play a game. Our result must be a subset of the edges of this graph. Notice that because after each game a team is eliminated, there cannot be a cycle in our subset so our subset is actually a tree (no cycles and every node), to get the maximum result just implement a maximum spanning tree algorithm on the graph.

For Silver 3, represent each team as a vertex. Then each edge represents a possible match, with the weight equal to the number of points scored. Note that every elimination tournament can be represented as a tree. The problem is now finding the maximal spanning tree.

Silver division HINT:

1st problem: KMP+stack; complexity=O(|S|+|T|); memory=O(|S|+|T|);

2nd problem: DP; complexity=O((K+R^2)*C); memory=O(R*C); (Alternative solution: DP; complexity=O(R*C); memory=O((K+R)*C))

3rd problem: MST+prim; complexity=O(N^2); memory=O(N);

I solved the 1st problem using a linked list to store the text, and deleted each instance of S.

const int MXN = (int)1e6 + 200;

you must change it to

const int MXN = (int)2e6 + 200;

my solution was O(nmlgm).

I used fenwick for every a[i][j]. (I push indices of columns that have at least on square have a[i][j]), then a dp solve problem.

maximum time is 300 ms.

here u go :) http://pastebin.com/bQvE4qFY

Your code is too awful because you wrote code by define.Primarily you will write code normal and ask again . Thank you for downvotes

horse > monkey

Upon deletion, you are going back S.length() spaces and researching. At the worst case it's still O(TS). You need to find a way to memoize so that you don't have to go back S.length() spaces.