POJ - 2533 Longest Ordered Subsequence

第一种情况，序列只有他自己 dp[i] = 1

第二种情况，a[i] > a[j]    i>j     dp[i] = max(dp[i],dp[j]+1)

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<string>
#include<map>
#define INF 0x3f3f3f3f
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define fori(x) for(int i=0;i<x;i++)
#define forj(x) for(int j=0;j<x;j++)
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(y))
#define sca(x) scanf("%d", &x)
#define scas(x) scanf("%s",x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define pris(x) printf("%s\n",x)
#define prl(x) printf("%lld\n",x)
//#include <bits/stdc++.h>

typedef long long ll;
const int maxn=1e6+7;
const int mod=1e9+7;
const double eps=1e-8;

using namespace std;

int dp[maxn];
int a[maxn];
int main()
{
    int n;
    int res;
    sca(n);
    res = -1;
    rep(i,0,n)
    {
      sca(a[i]);
    }
    rep(i,0,n)
    {
      dp[i] = 1;
       rep(j,0,i)
       {
         if(a[i] > a[j])
          {
            dp[i] = max(dp[i], dp[j] + 1);

          }
       }
       res = max(res,dp[i]);
    }
    pri(res);

    return 0;
}