CF10D LCIS（线性DP）

题意：\(LCIS\)输出方案

变迁の时刻，标记它
P.S:特判没\(LCIS\)的情况

//#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))

#define ON_DEBUGG

#ifdef ON_DEBUGG

#define D_e_Line printf("-----------\n")
#define D_e(x) std::cout << (#x) << " : " <<x << "\n"
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out2.txt", "w", stdout)
#define Pause() system("pause")
#include <ctime>
#define TIME() fprintf(stderr, "\nTIME :¡¡%.3lfms\n", clock() * 1000.0 / CLOCKS_PER_SEC)

#else

#define D_e_Line ;
#define D_e(x) ;
#define FileOpen() ;
#define FilSave ;
#define Pause() ;
#define TIME() ;

#endif

struct ios {
	template<typename ATP> ios& operator >> (ATP &x) {
		x = 0; int f = 1; char c;
		for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
		while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
		x *= f;
		return *this;
	}
}io;

using namespace std;

template<typename ATP> inline ATP Min(ATP a, ATP b) {
	return a < b ? a : b;
}
template<typename ATP> inline ATP Max(ATP a, ATP b) {
	return a > b ? a : b;
}
template<typename ATP> inline ATP Abs(ATP a) {
	return a < 0 ? -a : a;
}

const int N = 507;

#define int long long
int a[N], b[N], f[N][N];
int n, m;
pair<int, int> from[N][N];

inline void DFS(int x, int y) {
	if(!x) return;
	DFS(from[x][y].first, from[x][y].second);
	if(from[x][y].second != y) printf("%lld ", b[y]);
}
#undef int
int main() {
#define int long long
//FileOpen();
//freopen("okarin.in", "r", stdin);
//freopen("okarin.out", "w", stdout);
    io >> n;
    R(i,1,n) io >> a[i];
	io >> m;
	R(i,1,m) io >> b[i]; 
    R(i,1,n){
        int val=0;
        pair<int,int> p = make_pair(0, 0);
        R(j,1,m){
            if(a[i] == b[j]){
            	f[i][j] = val + 1;
				from[i][j] = p;
			}
            else{
            	f[i][j] = f[i - 1][j];
				from[i][j] = from[i - 1][j];
			}
            if(b[j] < a[i])
                if(val < f[i-1][j]){
                	p = make_pair(i - 1, j);
                	val = f[i - 1][j];
                }
        }
    }
    int now = 1;
    R(i,2,m){
    	if(f[n][i] > f[n][now]){
    		now = i;
		}
	}
    printf("%lld\n", f[n][now]);
    if(f[n][now]) DFS(n,now);
    return 0;
}