POJ-2533(最长上升子序列(简单dp))

题目名称:Longest Ordered Subsequence 题目链接:https://vjudge.net/problem/POJ-2533简单的动态规划, 状态转移方程为:dp[i] = max(dp[i], dp[j] + 1) 当 a[j] > a[i] && j < i;

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 1005;int a[MAXN], dp[MAXN];int main(){    int n;    while(scanf("%d", &n) != EOF)    {        for(int i = 0; i < n; ++i) scanf("%d", &a[i]);        //dp        int res = -1;        for(int i = 0; i < n; ++i)        {            dp[i] = 1;            for(int j = 0; j < i; ++j)                if(a[j] < a[i]) dp[i] = max(dp[i], dp[j]+1);            res = max(res, dp[i]);        }        printf("%d\n", res);    }    return 0;}