Codeforces Round #658 (Div. 2) A. Common Subsequence

A. Common Subsequence

题目链接-A. Common Subsequence


题目大意
给出两个数组，请你找出二者最短的公共子序列

解题思路

• 如果两个数组中没有一样的数字，那说明没有公共子序列，直接输出NO即可
• 如果有相同的数字，最短公共子序列肯定为$1$1，任意输出一个相同的数字即可
• 具体操作见代码

附上代码

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
#define int long long
#define lowbit(x) (x &(-x))
#define endl '\n'
using namespace std;
const int INF=0x3f3f3f3f;
const int dir[4][2]={-1,0,1,0,0,-1,0,1};
const double PI=acos(-1.0);
const double e=exp(1.0);
const double eps=1e-10;
const int M=1e9+7;
const int N=2e5+10;
typedef long long ll;
typedef pair<int,int> PII;
typedef unsigned long long ull;
int a[N],b[N];
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);

	int t;
	cin>>t;
	while(t--){
		int ass=0;
		map<int,int> mp;
		int n,m;
		cin>>n>>m;
		for(int i=1;i<=n;i++){
			cin>>a[i];
			mp[a[i]]=i;
		}
		for(int i=1;i<=m;i++){
			cin>>b[i];
			if(mp[b[i]]){
				ass=b[i];
			}
		}
		if(!ass) cout<<"NO"<<endl;
		else cout<<"YES"<<endl<<1<<" "<<ass<<endl;
	}
	return 0;
}