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.

 

题意：依旧是一道LIS的裸题

思路：要注意数据比较多，要用二分的优化法，原本是与POJ2533一样的题，但是用2533的代码只改下数组大小是过不了的，WA了几次，后来发现二分函数需要改进，于是改成了下面的方式，而且这个二分函数用在2533也是可以的

 

#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\n",LIS());
    }

    return 0;
}