Codeforces模擬賽，題解及體會

從某種角度來說，在一個穩定舉辦賽事的OJ上寫題的效率是比自己找題更高的。這一類的OJ包括且不限於CodeForces,Atcoder,Topcoder,Codechef,等等。此處放置部分模擬賽題解，不是給神看的，給人看吧。

Contest #402 Div.1

A. String Game:

Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.

Sergey gives Nastya the word t and wants to get the word p out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters’ indices of the word t: a1… a|t|. We denote the length of word x as |x|. Note that after removing one letter, the indices of other letters don’t change. For example, if t = “nastya” and a = [4, 1, 5, 3, 2, 6] then removals make the following sequence of words “nastya” “nastya” “nastya” “nastya” “nastya” “nastya” “nastya”.

Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word p. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.

It is guaranteed that the word p can be obtained by removing the letters from word t.

Input

The first and second lines of the input contain the words t and p, respectively. Words are composed of lowercase letters of the Latin alphabet (1 ≤ |p| < |t| ≤ 200 000). It is guaranteed that the word p can be obtained by removing the letters from word t.

Next line contains a permutation a1, a2, …, a|t| of letter indices that specifies the order in which Nastya removes letters of t (1 ≤ ai ≤ |t|, all ai are distinct).

題解：

直接二分答案，然後用p去匹配t中剩餘的字元即可。
複雜度 O(nlogn)

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B. Bitwise Formula

Bob recently read about bitwise operations used in computers: AND, OR and XOR. He have studied their properties and invented a new game.

Initially, Bob chooses integer m, bit depth of the game, which means that all numbers in the game will consist of m bits. Then he asks Peter to choose some m-bit number. After that, Bob computes the values of n variables. Each variable is assigned either a constant m-bit number or result of bitwise operation. Operands of the operation may be either variables defined before, or the number, chosen by Peter. After that, Peter’s score equals to the sum of all variable values.

Bob wants to know, what number Peter needs to choose to get the minimum possible score, and what number he needs to choose to get the maximum possible score. In both cases, if there are several ways to get the same score, find the minimum number, which he can choose.

Input

The first line contains two integers n and m, the number of variables and bit depth, respectively (1 ≤ n ≤ 5000; 1 ≤ m ≤ 1000).

The following n lines contain descriptions of the variables. Each line describes exactly one variable. Description has the following format: name of a new variable, space, sign “:=”, space, followed by one of:

Binary number of exactly m bits.
The first operand, space, bitwise operation (“AND”, “OR” or “XOR”), space, the second operand. Each operand is either the name of variable defined before or symbol ‘?’, indicating the number chosen by Peter.
Variable names are strings consisting of lowercase Latin letters with length at most 10. All variable names are different.

Output

In the first line output the minimum number that should be chosen by Peter, to make the sum of all variable values minimum possible, in the second line output the minimum number that should be chosen by Peter, to make the sum of all variable values maximum possible. Both numbers should be printed as m-bit binary numbers.

題解

從高到低，按位考慮此位選1還是0能使得所有變數在這一位上出現的1最多/少。程式碼難度較高。

C. Peterson Polyglot

Peterson loves to learn new languages, but his favorite hobby is making new ones. Language is a set of words, and word is a sequence of lowercase Latin letters.

Peterson makes new language every morning. It is difficult task to store the whole language, so Peterson have invented new data structure for storing his languages which is called broom. Broom is rooted tree with edges marked with letters. Initially broom is represented by the only vertex — the root of the broom. When Peterson wants to add new word to the language he stands at the root and processes the letters of new word one by one. Consider that Peterson stands at the vertex u. If there is an edge from u marked with current letter, Peterson goes through this edge. Otherwise Peterson adds new edge from u to the new vertex v, marks it with the current letter and goes through the new edge. Size of broom is the number of vertices in it.

In the evening after working day Peterson can’t understand the language he made this morning. It is too difficult for bored Peterson and he tries to make it simpler. Simplification of the language is the process of erasing some letters from some words of this language. Formally, Peterson takes some positive integer p and erases p-th letter from all the words of this language having length at least p. Letters in words are indexed starting by 1. Peterson considers that simplification should change at least one word, i.e. there has to be at least one word of length at least p. Peterson tries to make his language as simple as possible, so he wants to choose p such that the size of the broom for his simplified language is as small as possible.

Peterson is pretty annoyed with this task so he asks you for help. Write a program to find the smallest possible size of the broom and integer p.

Input

The first line of input contains integer n (2 ≤ n ≤ 3·105) — the size of the broom.

Next n - 1 lines describe the broom: i-th of them contains integers ui, vi and letter xi — describing the edge from ui to vi marked with letter xi.

Vertices are numbered from 1 to n. All xi are lowercase latin letters. Vertex 1 is the root of the broom.

Edges describe correct broom which is made from Peterson’s language.

Output

The first line of output should contain the minimum possible size of the broom after its simplification. The second line of output should contain integer p to choose. If there are several suitable p values, print the smallest one.

題解

對於每一層的每一個節點，直接啟發式合併計算可以省下的節點個數。
複雜度O(nlogn)

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Contest #345 Div.1

C. Table Compression

Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the new knowledge, Petya is now developing the new compression algorithm which he wants to name dis.

Petya decided to compress tables. He is given a table a consisting of n rows and m columns that is filled with positive integers. He wants to build the table a’ consisting of positive integers such that the relative order of the elements in each row and each column remains the same. That is, if in some row i of the initial table ai, j < ai, k, then in the resulting table a’i, j < a’i, k, and if ai, j = ai, k then a’i, j = a’i, k. Similarly, if in some column j of the initial table ai, j < ap, j then in compressed table a’i, j < a’p, j and if ai, j = ap, j then a’i, j = a’p, j.

Because large values require more space to store them, the maximum value in a’ should be as small as possible.

Petya is good in theory, however, he needs your help to implement the algorithm.

Input

The first line of the input contains two integers n and m (, the number of rows and the number of columns of the table respectively.

Each of the following n rows contain m integers ai, j (1 ≤ ai, j ≤ 109) that are the values in the table.

Output

Output the compressed table in form of n lines each containing m integers.

If there exist several answers such that the maximum number in the compressed table is minimum possible, you are allowed to output any of them.

題解

將所有數從小到大排序，每次處理大小相等的。
將相同行/列的數合併在並查集內，每次填行/列最大值+1。