Codeforces Round #738 (Div. 2) D1. Mocha and Diana (Easy Version)

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题意：

给你两个相同数量点的森林，问你最多能在两个森林中加多少相同的边。

思路：

两个循环即可搞定，用并查集判断加入边$<i,j>$后是否存在环。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<math.h>
using namespace std;
#define ll long long
#define mod 998244353ll
int f1[1010], f2[1010];
int find1(int x)
{
  return f1[x]==x? x : f1[x]=find1(f1[x]);
}

int find2(int x)
{
  return f2[x]==x? x : f2[x]=find2(f2[x]);
}
struct node{
  int x,y;
};
queue<node>q;

int main()
{
    int n,m1,m2;
    cin>>n>>m1>>m2;
    for(int i = 1; i <= n; i++)f1[i] = f2[i] = i;
    while(m1--)
    {
      int u,v;
      cin>>u>>v;
      f1[find1(u)] = find1(v);
    }
    while(m2--)
    {
      int u,v;
      cin>>u>>v;
      f2[find2(u)] = find2(v);
    }
    for(int i = 1; i <= n; i++)
    {
      for(int j = i+1; j <= n; j++)
      {
        if(find1(i) != find1(j) && find2(i) != find2(j))
        {
          node now;
          now.x = i,now.y = j;
          q.push(now);
          f1[find1(i)] = find1(j);
          f2[find2(i)] = find2(j);
        }
      }
    }
    cout<<q.size()<<endl;
    while(!q.empty())
    {
      node now = q.front();
      q.pop();
      cout<<now.x<<" "<<now.y<<endl;
    }
  }