Dynamic rules (18) - and check basic questions - gangs

There are N people living in a certain city , any two acquaintances are either friends or enemies, and satisfy:

    1. My   friend's friend is my friend;

    2. The enemy of my   enemy is my friend;

All people who are friends form a gang. Tell you M pieces of information about these N people , that is, certain two people are friends, or certain two people are enemies, please write a program to calculate how many gangs there may be in this city at most?

enter

The first line contains an integer N , the second line contains an integer M , 1<N<=1000,1<=M<=5000 ;

The next M lines describe M pieces of information, and the content is one of the following: " 0 XY " means that X and Y are friends; " 1 XY " means that X and Y are enemies.

output

Contains an integer, the maximum possible number of gangs.

sample input

6 4

1 1 4

0 3 5

0 4 6

1 1 2

sample output

3

#include <cstdio>
#include <iostream>
using namespace std;
int fa[1001], vis[1001];
int n, m, s, i, x, y, z, r1, r2;
int find(int x)
{
	return fa[x] == x ? x : fa[x] = find(fa[x]);
}
int main()
{
	cin >> n >> m;
	for (i = 1; i <= n; i++)
		fa[i] = i;
	for (i = 1; i <= m; i++)
	{
		cin >> z >> x >> y;
		if (z == 0)
		{
			r1 = find(x);
			r2 = find(y);
			fa[r2] = r1;
		}
		if (z == 1)
		{
			if (vis[x])
			{
				r1 = find(y);
				r2 = find(vis[x]);
				fa[r2] = r1;
			}
			else
				vis[x] = y;
			if (vis[y])
			{
				r1 = find(x);
				r2 = find(vis[y]);
				fa[r2] = r1;
			}
			else
				vis[y] = x;
		}
	}
	for (i = 1; i <= n; i++)
		if (fa[i] == i)
			s++;
	cout << s;
	return 0;
}