POJ 3903 Stock Exchange（LIS）
Stock ExchangeTime Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 4578
Accepted: 1618
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.
这题是很纯的LIS题，求最长单调递增子序列的长度。
1 #include<iostream>
2 #include<algorithm>
3 #include<cstdio>
4 #include<string.h>
5 using namespace std;
6 const int inf = 10001;
7 const int MAX = 100010;
8 int binary_search(int s[],int digit,int length)
9 {
10 int left=0,right=length,mid;
11 while(right!=left)
12 {
13 mid=(left+right)/2;
14 if(digit==s)
15 return mid;
16 else if(digit<s)
17 right=mid;
18 else
19 left=mid+1;
20 }
21 return left;
22 }
23 int main()
24 {
25 int n,j;
26 while(~scanf("%d",&n))
27 {
28 int a;
29 int s;
30 for(int i=1;i<=n;i++)
31 scanf("%d",&a);
32 s=a;
33 int len=1;
34 for(int i=1;i<=n;i++)
35 {
36 s=inf;
37 j=binary_search(s,a,len);
38 if(j==len)
39 len++;
40 s=a;
41 }
42 printf("%d\n",len-1);
43 }
44 return 0;
45 }
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