LIS-Program E

最大上升子序列

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
 
题目大意：给出一串序列，要求输出最大上升子序列。
 
分析：用动态规划，效率高；如果后一个数比前一个大，上升序列长度加1，否则长度减1，将后者较小的数代替前者。
 
代码如下：
 

#include <iostream>
#include <cstdio>
#include <algorithm>
const int maxn=100005;
using namespace std;
int a[maxn],dp[maxn],len,n;
int find(int x)
{
	int l=1,r=len,mid;
	while(l<=r)
	{
		mid=(l+r)>>1;
		if(x>dp[mid])
			l=mid+1;
		else
			r=mid-1;
	}
	return l;
}
int main()
{
	while(scanf("%d",&n)==1)
	{
		len=0;
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		for(int i=1;i<=n;i++)
		{
			int k=find(a[i]);
			dp[k]=a[i];
			len=max(len,k);
		}
		printf("%d\n",len);
	}
	return 0;
}