POJ 3903 Stock Exchange 最长上升子序列 模板题

Stock Exchange

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6314		Accepted: 2228

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

    题意：给定n个数，求取最长上升子序列的长度。

    分析：最长上升子序列模板题。dp[i]指的是以i为长度的上升子序列的末尾数的最小值。最长上升子序列理解可参见：http://blog.csdn.net/code_pang/article/details/8757380  AC代码：

#include
const int maxn=100005;
long long a[maxn];
int dp[maxn];//dp[i]指的是 i为长度的子序列的末尾数的最小值 
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		for(int i=0; i

    根据中间性质还可使用STL中的函数实现：见代码：

#include
#include
using namespace std;
const int maxn=100005;
long long a[maxn];
int dp[maxn];//dp[i]指的是 i为长度的子序列的末尾数的最小值
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		for(int i=0; i

    模板题需要好好理解掌握。POJ2533也为最长上升子序列的模板题。

    特记下，以备后日回顾。