一个连通块中的两个点之间不能添加边,所以想到使用并查集维护集合。设第一个人为 $A$,第二个人是 $B$,那么 $A$ 中的两个点 $x, y$ 能够相连的条件是 $B$ 中的 $x, y$ 不在一个连通块中。如果在 $x, y$ 之间添加边,此时 $A, B$ 中的连通块数量都减少 $1$,当 $B$ 的连通块数量是 $1$ 的时候,就不能再添加边了。对于 $x, y$ 来说,其实换成对应的连通块中的其它符合条件的两个点都可以,它们是等价的,最终效果都一样。最终局面是 $B$ 只有一个连通块,$A$ 的连通块至少有一个。因此检查 $A$ 中的所有点对,能连的就放到答案中即可。
// Date: Wed Jan 17 19:55:47 2024
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;
template <class T> using pq = priority_queue<T>;
template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353;
const ll INFL = 0x3f3f3f3f'3f3f3f3f;
const double eps = 1e-8;
const int dir[8][2] = {
{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};
const ull Pr = 131;
#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])
#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nemp(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())
#define NL cout << '\n';
template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }
template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
for (t &x : vec)
in >> x;
return in;
}
template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
int n = SZ(vec);
For(i, 0, n) {
out << vec[i];
if (i < n - 1)
out << ' ';
}
return out;
}
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V> void __print(const pair<T, V> &x) {
cerr << '{';
__print(x.first);
cerr << ", ";
__print(x.second);
cerr << '}';
}
template <typename T> void __print(const T &x) {
int f = 0;
cerr << '{';
for (auto &i : x)
cerr << (f++ ? ", " : ""), __print(i);
cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V> void _print(T t, V... v) {
__print(t);
if (sizeof...(v))
cerr << ", ";
_print(v...);
}
#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z) \
cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#define dbg(x...) \
cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \
_print(x); \
cerr << "\e[39m" << endl;
#else
#define debug1
#define debug2
#define debug3
#define dbg(x...)
#endif
const int N = 1100;
int f[2][N], sz[2][N], n, m1, m2;
void Init() {
For1(i, 1, n) {
f[0][i] = f[1][i] = i;
sz[0][i] = sz[1][i] = 1;
}
}
int Find(int i, int x) {
if (f[i][x] == x)
return x;
return f[i][x] = Find(i, f[i][x]);
}
void Union(int i, int x, int y) {
int rx = Find(i, x), ry = Find(i, y);
if (rx == ry)
return;
f[i][ry] = rx;
}
void solve() {
while (cin >> n >> m1 >> m2) {
int u, v;
Init();
For(i, 0, m1) {
cin >> u >> v;
Union(0, u, v);
}
For(i, 0, m2) {
cin >> u >> v;
Union(1, u, v);
}
vector<PII> res;
For1(i, 1, n) {
For1(j, i + 1, n) {
int ri = Find(0, i), rj = Find(0, j);
if (ri != rj) {
int ri1 = Find(1, i), rj1 = Find(1, j);
if (ri1 != rj1) {
res.pb({i, j});
Union(1, i, j);
Union(0, i, j);
}
}
}
}
int len = SZ(res);
cout << len << '\n';
if (len) {
for (auto &[x, y] : res) {
cout << x << ' ' << y << '\n';
}
}
}
}
int main(void) {
#ifdef _DEBUG
freopen("1559d1.in", "r", stdin);
#endif
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int T = 1;
while (T--) {
solve();
}
return 0;
}