Common Subsequence（贪心）

题目
题意：给定两个序列，求两者的最短公共子序列
思路：最短公共子序列，显然就是1了。

#include
using namespace std;
#define ll long long
const int maxn = 2010;

int a[maxn], b[maxn];
int mp[maxn], n, m;
int x;
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d%d",&n, &m);
        memset(mp, 0 ,sizeof(mp));
        for(int i = 1;i <= n; ++i) {
            scanf("%d", &a[i]);
            mp[a[i]] = 1;
        }
        for(int i = 1;i <= m; ++i) {
            scanf("%d", &b[i]);
        }
        bool flag = 0;
        int ans;
        for(int i = 1;i <= m; ++i) {
            if(mp[b[i]]) {
                ans = b[i];
                flag = 1;
                break;
            }
        }
        if(flag) {
            printf("YES\n1 %d\n",ans);
        } else {
            printf("NO\n");
        }
    }
    return 0;
}