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.
题目大意:求最大的上升子序列。
代码：
1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 const int maxn=100000+10; 5 int n,a[maxn],b[maxn],t; 6 int erfen(int i) 7 { 8 int l,r,m; 9 l=0,r=t-1; 10 while(l<=r) 11 { 12 m=l+(r-l)/2; 13 if(a[i]>b[m]) 14 l=m+1; 15 else r=m-1; 16 } 17 return l; 18 } 19 int main() 20 { 21 int i,j; 22 while(cin>>n) 23 { 24 for(i=0;i<n;i++) 25 { 26 cin>>a[i]; 27 } 28 t=0; 29 b[0]=a[0]; 30 for(i=1;i<n;i++) 31 { 32 if(a[i]>b[t]) 33 { 34 b[++t]=a[i]; 35 } 36 else 37 { 38 int d=erfen(i); 39 b[d]=a[i]; 40 } 41 42 } 43 cout<<t+1<<endl; 44 45 } 46 return 0; 47 }
这里使用的二分查找。
用2重循环会超时。