POJ 2533 Longest Ordered Subsequence(基础dp)

/*基础dpN - Longest Ordered Subsequence时间: 2017/02/23题意：题解：裸LIS*/#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<queue>#include<map>using namespace std;#define N 1010#define INF 0x3f3f3f3fint dp[N];int a[N];int main(){    int n;    while(~scanf("%d",&n))    {        memset(dp,0,sizeof(dp));        int mx = 0;        for(int i = 0; i < n; i++)        {            scanf("%d",&a[i]);            dp[i] = 1;            for(int j = i-1; j >= 0; j--)            {                if(a[i] > a[j])                    dp[i] = max(dp[j]+1,dp[i]);            }            mx = max(mx,dp[i]);        }        cout << mx << endl;    }    return 0;}