Poj 3903 Stock Exchange(LIS)

题目链接：http://poj.org/problem?id=3903

Stock Exchange

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6404 Accepted: 2275

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.

Source

Southeastern European Regional Programming Contest 2008

题目大意：    约翰每天都在追随股价上涨的趋势。给定一个数字序列P1，P2，…，Pn代表股票价格上升趋势是一个序列p1＜P2 <… 约翰的问题是很快找到最长的上升趋势。

解析：LIS，nLog(n)

代码如下：

#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<set>#include<string>#include<cstdio>#include<cstring>#include<cctype>#include<cmath>#define N 100009using namespace std;const int inf = 0x3f3f3f3f;const int mod = 1e9 + 7;const double eps = 1e-8;const double pi = acos(-1.0);typedef long long LL;int a[N], ans[N];int Seach(int l, int r, int k){    int m = (l + r) >> 1;    if(m == 0 && k > ans[m] && k < ans[1]) return 1;    if(m == 0) return 0;    if(ans[m] >= k && ans[m - 1] < k) return m;    if(k > ans[m]) return Seach(m + 1, r, k);    return Seach(l, m, k);}int main(){    int i, n;    while(~scanf("%d", &n))    {        for(i = 1; i <= n; i++)        {            scanf("%d", &a[i]);        }        ans[0] = a[1];        int r = 0;        for(i = 2; i <= n; i++)        {            if(ans[r] < a[i])            {                ans[++r] = a[i]; continue;            }            else if(ans[r] == a[i]) continue;            int k = Seach(0, r, a[i]);            ans[k] = a[i];        }        printf("%d\n", r + 1);    }    return 0;}