[2078] Luo Gu friend

Topic background

In Xiaoming Company A, Company B red work.

Title Description

Both employees have a characteristic: a company's employees are homosexual.

A company has N employees, of which P for friends. Company B has M employees, including Q for friends. Certain friends or friends of friends.

Each pair consisting of a friend relationship with two integers (Xi, Yi), represents a number of friends were Xi, Yi. Men's number is positive, the woman's number is negative. Bob is the number 1, red number is -1.

As we all know, Xiao Ming and red are friends, then you write a program between the two companies calculated, and the little red man by Xiao Ming up to a total understanding of how much the couple can be formulated. (Including themselves)

Input Format

Line 1, four spaces separated by a positive integer N, M, P, Q.

After the P lines of two positive integers Xi, Yi.

After the Q lines of two negative integers Xi, Yi.

Output Format

Line, a positive integer representing the people by Bob and Alice know how much the couple can be up to a total of dubbed. (Including themselves)

Sample input and output

Input # 1

Output # 1

Description / Tips

Data for 30%, N, M <= 100, P, Q <= 200

Data for 80%, N, M <= 4000, P, Q <= 10000.

For all data, N, M <= 10000, P, Q <= 20000.

Solution: How many boys will seek understanding and 1 girls know how many people -1, then put it all together

Whichever is less direct access to. (After all, not polygamy hhh)

#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cstdio>
using namespace std;

const int N=20005;
int n,m,p,q,x,y,a1,a2;
int fa1[N],fa2[N],ans1,ans2;

int find1(int x){
   if(fa1[x]==x) return x;
   return fa1[x]=find1(fa1[x]);
}
int find2(int x){
   if(fa2[x]==x) return x;
   return fa2[x]=find2(fa2[x]);
}

int main(){
   
    scanf("%d %d %d %d",&n,&m,&p,&q);
    
    for(int i=1;i<=n;i++) fa1[i]=i;
    for(int i=1;i<=m;i++) fa2[i]=i;
    
    for(int i=1;i<=p;i++){
        scanf("%d %d",&x,&y);
        a1=find1(x); a2=find1(y);
        if(a1!=a2) fa1[a2]=a1;
    }
    for(int i=1;i<=q;i++){
        scanf("%d %d",&x,&y);
        x=-x; y=-y;
        a1=find2(x); a2=find2(y);
        if(a1!=a2) fa2[a2]=a1;
    }
    
    for(int i=1;i<=n;i++)
        if(find1(i)==find1(1)) ans1++;
    for(int i=1;i<=m;i++)
        if(find2(i)==find2(1)) ans2++;
        
    printf("%d",min(ans1,ans2));
    return 0;
}