[POJ2533]Longest Ordered Subsequence

題目連接：http://poj.org/problem?id=2533

描述：

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.

翻譯：

找一串數字最長上升子序列的數字個數。（手動狗頭）

沒有難度的一個dp題，範圍不算大，最普通的方法就能夠過了

想優化看到了其餘博主說的二分，不過我沒有用

等複習到二分再來試試吧

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

int n,a[10005],dp[10005],ans;

int main(){
    n=read();
    for(int i=1;i<=n;i++)
        a[i]=read();
    for(int i=2;i<=n;i++)
        for(int j=1;j<i;j++)
            if(a[i]>a[j])dp[i]=max(dp[i],dp[j]+1);
    for(int i=1;i<=n;i++)
        ans=max(ans,dp[i]);
    cout<<ans+1;
    return 0;
}

View Code