Food chain (disjoint-set classic title)

Ideas: Write this question https://www.cnblogs.com/2462478392Lee/p/11343747.html it is easy to understand, and came up with the idea. Open a 3 * n array, the x and y kind, predators, and prey do extended domain,

Then using the relationship which do disjoint-set, each case has three. When x and y are the same, the same kind prey, predator predator of x and y, prey and y x. When the prey x y, predator x and y, x of prey

And predator-y y, x of prey. After establishing and check to see how many sets do not meet the current situation and check the set you can find the answer.

 

#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<cstdio>
#include<cmath>
#define ll long long
using namespace std;
int fa[2000000];
int get(int k)
{ 
    return fa[k]==k?k:fa[k]=get(fa[k]); 
}
void merge(int x,int y)
{
    fa[get(x)]=get(y); 
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=3*n;i++)
    {
        fa[i]=i;
    }
    int ans=0;
    for(int i=1;i<=m;i++)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        if(b>n||c>n)
        ans++;
        else if(a==1)
        {        
            if(get(b)==get(c+n)||get(b)==get(c+n+n))
            ans++;
            else
            {
                merge(b,c);
                merge(b+n,c+n);
                merge(b+n+n,c+n+n);
            }
        }
        else
        {
            if(b==c||get(b)==get(c)||get(c+n)==get(b))
            ans++;
            else
            {
                merge(b+n,c);
                merge(b,c+n+n);
                merge(b+n+n,c+n);
            }
        }
    }
    printf("%d\n",ans);
}