POJ 3903 Stock Exchange(树状数组求最长上升子序列)

最新推荐文章于 2021-03-09 11:37:19 发布

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本文介绍了一种用于寻找股票价格中最长上升趋势的高效算法。该算法通过预处理和排序技术来加速查找过程，并使用线段树结构进行快速查询与更新，以解决大规模数据集上的性能瓶颈。

Stock Exchange

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:8000		Accepted: 2775

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.

#include<cstdio>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=1e5+7;
int a[maxn],val[maxn],dp[maxn],n;
int query(int i)
{
    int s=0;
    while(i>0)
    {
        s=max(dp[i],s);
        i-=i&-i;
    }
    return s;
}
void add(int i,int x)
{
    while(i<=n)
    {
        dp[i]=max(dp[i],x);
        i+=i&-i;
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)scanf("%d",&a[i]),val[i]=a[i];
        sort(a,a+n);
        int len=unique(a,a+n)-a;
        memset(dp,0,sizeof(dp));
        int ans=0,tmp;
        for(int i=0;i<n;i++)
        {
            int id=lower_bound(a,a+len,val[i])-a+1;
            tmp=query(id-1)+1;
            ans=max(ans,tmp);
            add(id,tmp);
        }
        printf("%d\n",ans);
    }
    return 0;
}