poj 3903 Stock Exchange

Description

The world financial crisis is quite a subject. Some people are more relaxed while others are quite anxious. John is one of them. He is very concerned about the evolution of the stock exchange. He follows stock prices every day looking for rising trends. Given a sequence of numbers p1, p2,...,pn representing stock prices, a rising trend is a subsequence pi1 < pi2 < ... < pik, with i1 < i2 < ... < ik. John’s problem is to find very quickly the longest rising trend.

Input

Each data set in the file stands for a particular set of stock prices. A data set starts with the length L (L ≤ 100000) of the sequence of numbers, followed by the numbers (a number fits a long integer).
White spaces can occur freely in the input. The input data are correct and terminate with an end of file.

Output

The program prints the length of the longest rising trend.
For each set of data the program prints the result to the standard output from the beginning of a line.

Sample Input

6 
5 2 1 4 5 3 
3  
1 1 1 
4 
4 3 2 1

Sample Output

3 
1 
1

Hint

There are three data sets. In the first case, the length L of the sequence is 6. The sequence is 5, 2, 1, 4, 5, 3. The result for the data set is the length of the longest rising trend: 3.

Source

Southeastern European Regional Programming Contest 2008

 

题目大意就是求一个数列的最长上升子序列。

 

本来想只要一个导弹就可以了，可是看下范围，裸的n方肯定爆。有回忆这写了一下nlogn的优化，就过了。

具体的就是开一个数组s(程序中用的是f)，s[i]记录长度为i的 合法子序列中 最后一个位最小的值。

我们顺序处理原序列中的每一个值，同时每次维护s数组。

这样对于原序列中的每一个值a[i]，就是到s数组中查找小于a[i]的最大的s[j]，然后把s[j+1]赋成a[i]，这样就能保证s数组的合法性。

我们可以很容易发现，s数组是个严格递增的序列，这样，在查找上面提到的s[j]的时候就可以用二分，在最坏log n的时间就可以找到。在实际情况中，这样的时效是很优的。

 

AC CODE

 

program pku_3903;
var f:array[0..100000] of longint;
//============================================================================
procedure work;
var n,len,l,r,x,i,mid:longint;
begin
  readln(n); len:=0;
  for i:=1 to n do
  begin
      read(x);
      l:=0; r:=len;
      while l<=r do
      begin
          mid:=(l+r) shr 1;
          if f[mid]>=x then r:=mid-1 else l:=mid+1;
      end; f[l]:=x;      //为什么直接用左端点呢？我也不知道。。。观察就是这样的就用了。。。
      if l=len+1 then len:=l;
  end; writeln(len);
end;
//============================================================================
begin
  f[0]:=-maxlongint;
  while not(seekeof) do work;
end.