就是多了一步合并敌人的敌人，别的基本就是板子 建议训练: tzhsojor洛谷连接(这两是一道题

#include <bits/stdc++.h>
#define int long long int

using namespace std;

const int N = 2e5 + 7;
int n,m;
int fa[N << 1],e[N];

inline int read(){
	int x = 0,f = 1;
	char c = getchar();
	for(;!isdigit(c);c = getchar())
		if(c == '-')f = -1;
	for(;isdigit(c);c = getchar())
		x = (x << 1) + (x << 3) + c - '0';
	return x * f;
}

int get(int x){
	if(fa[x] == x)return x;
	return fa[x] = get(fa[x]);
}

void merge(int x,int y){
	x = get(x);
	y = get(y);
	fa[x] = y;
}

void print(int x,int y){
	if(get(x) == get(y)) printf("Y\n");
	else printf("N\n");
}

signed main(){
	n = read();m = read();
	
	for(int i = 1;i <= n;i ++){
		fa[i] = i;
		fa[i + n] = i + n;
	}
	
	for(int i = 1;i <= m;i ++){
		int p = read(),x = read(),y = read();
		if(!p) merge(x,y);
		else{
			if(!e[x]) e[x] = y;
			else{
				merge(e[x],y);
			}
			if(!e[y]) e[y] = x;
			else{
				merge(e[y],x);
			}
		}
	}
	
	int tot = 0;
	
	for(int i = 1;i <= n;i ++) if(fa[i] == i) tot++;
	printf("%lld",tot);
	
	return 0;
}