LIS O(n^2)模板

dp[j] = max(dp[k]+1,dp[j])(0<k<j&a[j]>a[k])

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;

const int inf = (1<<31)-1;
const int MAXN = 5e5+10;

int dp[MAXN];
int a[MAXN];

int main()
{
    int n;
    int mmax;
    while(~scanf("%d",&n)){
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        dp[1] = 1;
        mmax = 1;
        for(int i=2;i<=n;i++){
            for(int j=1;j<i;j++){
                if(a[i]>a[j]){
                    dp[i] = max(dp[j]+1,dp[i]);
                }
            }
            mmax = max(mmax,dp[i]);
        }
        cout<<mmax<<endl;
    }
    return 0;
}

 其实还有一种O(n*lgn)+O(n^2)的算法，先把原序列排序,再寻找LCS