poj 2533 Longest Ordered Subsequence （最长不下降子序列）

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

题目大意：就是让你求最长的不下降子序列

题目分析：很简单的一道dp，只是为了巩固一下自己对dp的理解，所以又做了一下，状态转移方程就是

  if（a[j]>a[i]）  dp[j]=max（dp[i]+1，dp[j]）；

  不过这道题目还有二分的写法，好像是省时间的，我

#include <cstdio>#include <cstring>#include <iostream>using namespace std;#define maxn 10005int dp[maxn],a[maxn];int main(){int n;while((scanf("%d",&n))!=EOF){memset(dp,0,sizeof(dp));memset(a,0,sizeof(a));int ans=0;for(int i=1;i<=n;i++)scanf("%d",&a[i]);    for(int i=1;i<=n;i++){    for(int j=i+1;j<=n;j++){    if(a[j]>a[i]&&dp[i]+1>dp[j])      dp[j]=dp[i]+1;}if(dp[i]>ans) ans=dp[i];}printf("%d\n",ans+1);}return 0;} 

然后，还有一种是省时间的二分法求最长不下降子序列方法。

就是开一个数组用来存储当前最长长度为k的不下降子序列的末尾元素，每次读入数据，如果大于f【k】，则更新k，k++，否则就把这个数据放在大于这个元素的最小f【i】。然后查找方法用二分，所以省时间。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;#define maxn 1005int f[maxn];int search(int a,int l,int r){int m;while(l<=r){m=(l+r)/2;if(f[m] == a){ l=m; return l;}  else if(f[m] > a)r=m-1;else l=m+1;}return l;//确保返回的是大于a的最小值 }int main(){int n,t,a;while((scanf("%d",&n))!=EOF){memset(f,0,sizeof(f));t=0;for(int i=1;i<=n;i++){scanf("%d",&a);int k=search(a,1,t);if(k<=t)  f[k]=a;else {  t=t+1;  f[t]=a;}} printf("%d\n",t);}return 0; }