ch_5101 LCIS

f[i][j]表示用a[1]-a[i]与b[1]-b[j]并且用了b[j]构成的LCIS，

优化：决策集合只增不减，记录最优就行，n^3 -> n^2

读入不知道为啥只能cin，快读要炸第四个点

#include<iostream>
#include<cstdio>

#define ri register int
#define u int

namespace opt {

    inline u in() {
        u x(0),f(1);
        char s(getchar());
        while(s<'0'||s>'9') {
            if(s=='-') f=-1;
            s=getchar();
        }
        while(s>='0'&&s<='9') {
            x=(x<<1)+(x<<3)+s-'0';
            s=getchar();
        }
        return x*f;
    }

}

using opt::in;

#define NN 3005

namespace mainstay {
    
    using std::cin;

    u N,a[NN],b[NN],f[NN][NN];

    inline void solve() {
        cin>>N;
        for(ri i(1); i<=N; ++i) cin>>a[i];
        for(ri i(1); i<=N; ++i) cin>>b[i];
        for(ri i(1); i<=N; ++i) {
            u v(0);
            for(ri j(1); j<=N; ++j) {
                if(b[j]<a[i]) v=std::max(v,f[i-1][j]);
                if(a[i]^b[j]) f[i][j]=f[i-1][j];
                else f[i][j]=v+1;
            }
        }
        u ans(0);
        for(ri i(1); i<=N; ++i) ans=std::max(ans,f[N][i]);
        printf("%d",ans);
    }

}

int main() {

    //freopen("x.txt","r",stdin);
    std::ios::sync_with_stdio(false);
    mainstay::solve();

}