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

传送门

题意： 找到a，b序列的最短公共子序列。若没有输出NO，反之输出YES后再输出子序列的长度和子序列。

思路： 要求找最短，那直接找一个公共元素就好。

代码实现：

#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int t, n, m;
int a[N], b[N], vis[N];

signed main()
{
    IOS;

    cin >> t;
    while(t --){
       cin >> n >> m;
       me(vis);
       for(int i = 1; i <= n; i ++){
            cin >> a[i];
            vis[a[i]] = 1;
       }int ok = 0, ans = 0;
       for(int i = 1; i <= m; i ++){
            cin >> b[i];
            if(vis[b[i]] && !ok) ok = 1, ans = b[i];
       }
       if(ok) cout << "YES" << endl << 1 << " " << ans << endl;
       else cout << "NO" << endl;
    }

    return 0;
}