C++ :

//用0，1，2分别表示3种动物
#include<stdio.h>
#include<string.h>
#define Max 50010
//rank[i]:记层数; father[i]:记i的根结点; link[i]: 以i的根为0为基准i属于的动物种类为link[i];
int rank[Max];
//i从link[i]变为j种动物: link[i] = (link[i]+step[j][link[i]])%3
int step[3][3] = {{0, 2, 1}, {1, 0, 2}, {2, 1, 0}};
int n, k, d, x, y, wrong; 
struct PP
{
	int father; 
	int link;
}p[Max];
//初始化：每个元素就是一个集合
void set(int n)
{
	for(int i = 1; i <= n; i++)
	{ p[i].father = i; rank[i] = 0; p[i].link = 0; }
}
//寻找元素所在集合的根，使用路径压缩来优化该函数
PP f(int i)
{
	PP t;
	if(p[i].father != i) 
	{ 
		t = f(p[i].father);
		p[i].father = t.father;
		p[i].link = (t.link+p[i].link)%3;
	}
	return p[i];
}
// 若为真将x和y所在的集合合并
void Union(int a, int b)
{
	if(a != b){ //要合并的
		if(rank[a] >= rank[b]){ 
			p[b].father = a;
			if(rank[a] == rank[b]) rank[a]++;
			if(d == 2) p[b].link = (p[b].link+step[(p[x].link+1)%3][p[y].link])%3;
			else if(d == 1) p[b].link = (p[b].link+step[p[x].link][p[y].link])%3;
			
		}
		else if(rank[a] < rank[b]){
			p[a].father = b;
			if(d == 2) p[a].link = (p[a].link+step[(p[y].link+2)%3][p[x].link])%3;
			else if(d == 1) p[a].link = (p[a].link+step[p[y].link][p[x].link])%3;  
		}
	}
	else{ //不需要合并的
		if(d == 1 && p[x].link != p[y].link)wrong++; 
		else if(d == 2 && (p[x].link+1)%3 != p[y].link)wrong++; 
	}
}
int main()
{
	//freopen("eat.in", "r", stdin);
	//freopen("eat.out", "w", stdout);
	int i;
	PP t1, t2;
	scanf("%d%d", &n, &k);
	set(n);
	for(wrong = i = 0; i < k; i++)
	{
		scanf("%d%d%d", &d, &x, &y);
		if(x > n || y > n){ wrong++; continue; }
		if(x == y) {
			if(d == 2) wrong++;
			continue;
		}
		t1 = f(x); t2 = f(y); 
		Union(t1.father, t2.father);
	}
	printf("%d\n", wrong);
	return 0;
}

Pascal :

var
  i,x,y,k,n,ans:longint;
  t:integer;
  dad:array[0..50001] of word;
  r:array[0..50001] of byte;
function find(i:word):word;
var
  temp:longint;
begin
  if dad[i]=i then exit(i);
  temp:=dad[i];
  dad[i]:=find(dad[i]);
  r[i]:=(r[temp]+r[i]) mod 3;
  exit(dad[i]);
end;
procedure union(x,y,h:longint);
var
  a,b:longint;
begin
  a:=find(x);
  b:=find(y);
  dad[a]:=b;
  r[a]:=(r[y]+h-r[x]+3) mod 3;
end;
begin
  readln(n,k);
  for i:=1 to n do dad[i]:=i;
  for i:=1 to k do  begin
    readln(t,x,y);
   if (x>n) or (y>n) then begin inc(ans); continue; end;
    if t=1 then
     if find(x)=find(y) then
      begin
        if r[x]<>r[y] then begin inc(ans); continue; end;
      end
      else union(x,y,0);
    if t=2 then
      begin
        if x=y then
        begin
          inc(ans);
          continue;
        end;
        if find(x)=find(y) then
        begin
          if r[x]<>(r[y]+1) mod 3 then begin inc(ans); continue; end;
        end
        else union(x,y,1);
      end;
  end;
  writeln(ans);
end.