证券交易所 Stock Exchange 最长上升子序列 poj 3903

Stock Exchange

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3585		Accepted: 1296

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.

#include
#include
using namespace std;
int D[100001];
int binarysearch(int low,int high,int m)//二分查找
{
	int mid;
	mid=(low+high)/2;
	while(low<=high)
	{
		if(D[mid]=m)
			return mid;
		else
			if(D[mid]>n)
	{
		for(i=1;i<=n;++i)
			cin>>a[i];
		D[1]=a[1];
		len=1;
		for(i=2;i<=n;i++)
		{
			if(a[i]>D[len])
			{
				len++;
				D[len]=a[i];
			}
			else
			{
				j=binarysearch(1,len,a[i]);
				k=j+1;
				D[k]=a[i];
			} 
		}  
	cout<