poj2533

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 51188 Accepted: 22754

Description

A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence (a₁, a₂, ..., a_N) be any sequence (a_i1, a_i2, ..., a_iK), where 1 <= i₁ < i₂ < ... < i_K <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

71 7 3 5 9 4 8

Sample Output

4

就是求最长上升子序列的问题

这个问题我采用的是o（nlogn）的二分插值的方法做的

当然数据量不大，可以采用普通的方法做出来

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<algorithm>using namespace std;int st[1005];int binary(int num,int len){int low,high;low=1;high=len;int mid;while(low<high){mid=(low+high)/2;if(num>=st[mid]){low=mid+1;}else{high=mid;}}return low;}int main(){int a[1005];int n;int i,j;int top;int pos;while(~scanf("%d",&n)){for(i=1;i<=n;i++){scanf("%d",&a[i]);}st[1]=a[1];top=1;for(i=2;i<=n;i++){//printf("\n");if(a[i]>st[top]){top++;st[top]=a[i];}else{pos=binary(a[i],top);st[pos]=a[i];}}printf("%d\n",top);}    return 0;}