F - Two Exam(AtCoder Beginner Contest 238)

First of all, in accordance with the $P$ Attribute sort owner , according to $P$ Select people in ascending order of attributes , You only need to consider $Q$ Property size relationship .
set up $DP$ Status as $dp[i][j][k]$
Before presentation $i$ Personal choice $j$ individual , The smallest of them has not been selected $Q$ The attribute is $k$ The number of solutions
Consider the $i+1$ Choose or not
$$\begin{gathered} Such\; asfruitk>The\; firstiindividualpeopleOfQBelong\; tosex \\ dp[i][j+1][k]=(dp[i][j+1][k]+dp[i-1][j][k]) \\ nobe \\ dp[i][j][min(k,The\; firstiindividualpeopleOfQBelong\; tosex)]=(dp[i][j][min(k,The\; firstiindividualpeopleOfQBelong\; tosex)]+dp[i-1][j][k]) \end{gathered}$$

#include<bits/stdc++.h>
#define ll long long
using namespace std;
struct Stu
{
    
    int p,q;
    bool operator <(const Stu &ths)const
    {
    
        return p<ths.p;
    }
}stu[305];
int dp[305][305][305];// front i individual , Yes j individual , The smallest one not selected is k Number of alternatives 
const int mod=998244353;
int main()
{
    
    int n,K;
    scanf("%d%d",&n,&K);
    for(int i=1;i<=n;i++)scanf("%d",&stu[i].p);
    for(int i=1;i<=n;i++)scanf("%d",&stu[i].q);
    sort(stu+1,stu+1+n);
    dp[0][0][n+1]=1;
    for(int i=1;i<=n;i++)
    {
    
        for(int j=0;j<=i-1;j++)
        {
    
            for(int k=1;k<=n+1;k++)
            {
    
                dp[i][j][min(k,stu[i].q)]=(dp[i][j][min(k,stu[i].q)]+dp[i-1][j][k])%mod;
                if(k>stu[i].q)dp[i][j+1][k]=(dp[i][j+1][k]+dp[i-1][j][k])%mod;
            }

        }
// for(int j=0;j<=i;j++)
// for(int k=1;k<=n+1;k++)printf("dp[%d][%d][%d]=%d\n",i,j,k,dp[i][j][k]);
    }
    ll ans=0;
    for(int i=1;i<=301;i++)ans=(ans+dp[n][K][i])%mod;
    printf("%lld",ans);
    return 0;
}