poj 3903

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.

题意：

给定L个整数A1,A2，...，An，按照从左到右的顺序选出尽量多的整数，组成一个上升序列（子序列可以理解为：删除0个或者多个数，其他的数的吮吸不变）。例如，1，6，2，3，7，5，可以选出上升子序列1，2，3，5，也可以选出1，6，7，但前者更长，选出的上升子序列中相邻元素不能相等。

思路：

开辟一个栈，每次取栈顶元素s和读到的元素a做比较，如果a>s，  则加入栈；如果a<s，则二分查找栈中的比a大的第1个数，并替换。  最后序列长度为栈的长度。 

代码：

#include<cstdio>
using namespace std;
int const size=100010;
int stack[size];
int main()
{
    int L;
    int top,temp;
    while(scanf("%d",&L)!=EOF)
    {
        top=0;
        stack[0]=-1;//第一个数可能为0
        for(int i=0;i<L;i++)
        {
            scanf("%d",&temp);
            if(temp>stack[top])//比栈顶大的元素就入栈
            {
                stack[++top]=temp;
            }
            else
            {
                int left=1,right=top;
                int mid;
                while(left<=right)//二分检索栈中第一个比temp大的元素
                {
                    mid=(left+right)/2;
                    if(temp>stack[mid])
                        left=mid+1;
                    else
                        right=mid-1;
                }
                stack[left]=temp;
            }
        }
        printf("%d\n",top);
    }
    return 0;
}