Title Description

I lived in a city where n people, any two people who know is the enemy not friends, but also to meet:

1, my friend's friend is my friend;

2, the enemy of my enemy is my friend;

All are friends of people of a gang. Tell you this n m pieces of information about an individual that is a friend of two people, two people or a certain enemy, you write a program to calculate the maximum number of cities may have a gang?

Input Format

The first line n and m, 1 <n <1000,1≤m≤100000;

The m rows, each behavior p, x, y, p is 0 or 1, p is 0, x and y are represented by friends, p is 1, x and y are represented by the enemy.

Output Format

An integer representing the n individual may have up to several gangs.

SAMPLE INPUT

6 4

1 1 4

0 3 5

0 4 6

1 1 2

Sample Output

answer

If the input is between two friends, it would direct the merger; otherwise we will not able to save the first alas, have finished entering, then merge all the enemies to everyone.

#include <iostream>
#include <cstdio>

#define MAX_N 1000
#define MAX_M 100000

using namespace std;

int n;
int m;
int h[MAX_N + 5], p[MAX_M + MAX_M + 5], t[MAX_M + MAX_M + 5];
int r[MAX_N + 5];
int c[MAX_N + 5];
int ans;

int Root(int x)
{
    int R = x, tmp;
    while(R != r[R]) R = r[R];
    while(x != r[x]) tmp = r[x], r[x] = R, x = tmp;
    return R;
}

int main()
{
    scanf("%d%d", &n, &m);
    for(register int i = 1; i <= n; ++i)
    {
        r[i] = i;
    }
    int f, x, y;
    for(register int i = 1; i <= m; ++i)
    {
        scanf("%d%d%d", &f, &x, &y);
        if(f)
        {
            t[i] = y;
            p[i] = h[x];
            h[x] = i;
            t[i + m] = x;
            p[i + m] = h[y];
            h[y] = i + m;
        }
        else
        {
            if(Root(x) != Root(y))
            {
                r[Root(y)] = Root(x);
            }
        }
    }
    for(register int i = 1; i <= n; ++i)
    {
        x = t[h[i]];
        for(register int j = h[i]; j; j = p[j])
        {
            if(Root(x) != Root(t[j]))
            {
                r[Root(t[j])] = Root(x);
            }
        }
    }
    for(register int i = 1; i <= n; ++i)
    {
        c[Root(i)] = 1;
    }    
    for(register int i = 1; i <= n; ++i)
    {
        ans += c[i];
    }
    printf("%d", ans);
    return 0;
}

Reference program

[Solution] gang problem