POJ3903:Stock Exchange(LIS)

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.
 
제목: 여전히 LIS의 누드 문제
사고방식: 데이터가 비교적 많다는 것을 주의해야 한다. 이분의 최적화법을 사용해야 한다. 원래는 POJ2533과 같은 문제였지만 2533의 코드로 하수조의 크기만 바꾸는 것은 지나칠 수 없다. WA를 몇 번 했는데 나중에 이분함수가 개선되어야 한다는 것을 발견하여 아래의 방식으로 바꾸었다. 그리고 이 이분함수는 2533에 사용해도 된다.
 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int a[100005],c[100005],n;

int bin(int size,int k)
{
    int l = 1,r = size;
    while(l<=r)
    {
        int mid = (l+r)/2;
        if(k>c[mid])
            l = mid+1;
        else
            r = mid-1;
    }
    return l;
}

int LIS()
{
    int i,j,cnt=0,k;
    for(i = 1; i<=n; i++)
    {
        if(cnt == 0 || a[i]>c[cnt])
           c[++cnt] = a[i];
        else
        {
           k = bin(cnt,a[i]);
           c[k] = a[i];
        }
    }
    return cnt;
}

int main()
{
    long i;
    while(~scanf("%d",&n))
    {
        for(i = 1; i<=n; i++)
            scanf("%d",&a[i]);
        printf("%d
",LIS());
    }

    return 0;
}