poj3903 Stock Exchange（最长上升子序列）

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题目链接：

id=3903">http://poj.org/problem?id=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

第一种：http://blog.csdn.net/u012860063/article/details/34086819

代码例如以下：

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

int a[100047],dp[100047],n;

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

int LIS(int *a)
{
    int i,j,ans=1;
    dp[1] = a[1];
    for(i = 2; i<=n; i++)
    {
        if(a[i]<=dp[1])
            j = 1;
        else if(a[i]>dp[ans])
            j = ++ans;
        else
            j = bin(ans,a[i]);
        dp[j] = a[i];
    }
    return ans;
}
int main()
{
	int i;
	while(~scanf("%d",&n))
	{
		for(i = 1; i <= n; i++)
		{
			scanf("%d",&a[i]);
		}
		int ans = LIS(a);
		printf("%d\n",ans);
	}
	return 0;
}

另外一种使用了：lower_bound

代码例如以下：

#include<cstdio>
#include<algorithm>
using namespace std;
#define INF 0x3fffffff

int a[100047],dp[100047];

int main()
{
	int t,i;
	while(scanf("%d",&t)!=EOF)
	{
		for(i = 0; i <= t; i++)
			dp[i]=INF;
		for(i = 0; i < t; i++)
		{
			scanf("%d",&a[i]);
		}
		for(i = 0; i < t; i++)
		{
			*lower_bound(dp,dp+t,a[i])=a[i];
		}
		printf("%d\n",lower_bound(dp,dp+t,INF)-dp);
	}
	return 0;
}

附上解释的图片一张：