POJ 3903 Stock Exchange

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 

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.

程序分析：此题的意思就是找到最长递增子序列。但是值得注意的是如果这题由于数据较大，就不能采用常规的DP方法来解答，那样时间复杂度为O(n^2)，应采用二分，时间复杂度为n * logn。

程序代码：

#include <cstdio>   
#include<iostream>
#include <cstring>   
#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,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;  
}