240. food chain (disjoint-set + number theory)

There are three types of animals in the animal kingdom A, B, C, three types of animal food chain constitute an interesting ring.

A food B, B eat C, C eat A.

Animals prior N, number to 1-N.

Each animal is A, B, C in kind, but we do not know it in the end is what kind of.

Some people describe this relationship N animal food chain formed by two different ways:

The first argument is "1 XY", represents the X and Y are similar.

The second argument is "2 XY", X represents eat Y.

This person for N animals, with the above two statements, one sentence by sentence to say K, K this sentence some true, some false.

When one of the following three words, this sentence is a lie, the truth is otherwise.

1) if the current true, then some of the previous conflicts, is lie;
2), then the current X or Y is larger than N, is lie;
3) X represents eat, then the current X, is lie.

Your task is given according to the total number of N and K words, the output of lies.

Input Format

The first line of two integers N and K, separated by a space.

K The following three lines each is a positive integer D, X, Y, separated by a space between the two numbers, where D indicates the type of argument.

If D = 1, it indicates that X and Y are similar.

If D = 2, then X represents eat Y.

Output Format

Only one integer representing the number of lies.

data range

1≤N≤500001≤N≤50000,
0≤K≤1000000≤K≤100000

Sample input:

100 7
1 101 1 
2 1 2
2 2 3 
2 3 3 
1 1 3 
2 3 1 
1 5 5

Sample output:

3

 Thinking: According to the root of the distance modulus value classification 3

           

 

        The relationship between eating and being eaten three animals are built up

          

 

 

        Therefore, the use disjoint-set distance from the node to the root node maintains

         FIG give a d [px] to a reasonable value

     　

 

          Figure II to d [px] to a reasonable value

　　 

 

 Code:

import java.util.*;

public class Main{
    static final int N=100005;
    static int p[]=new int[N];
    static int d[]=new int[N];
    static int n,k;
    static int find(int x){
        if(p[x]!=x){
            int t=find(p[x]);
            d[x]+=d[p[x]];
            p[x]=t;
        }
        return p[x];
    }
    
    public static void main(String []args){
        Scanner scan=new Scanner(System.in);
        n=scan.nextInt();
        k=scan.nextInt();
        for(int i=1;i<=n;i++) p[i]=i;
        int res=0;
        while(k-->0){
            int D=scan.nextInt();
            int x=scan.nextInt();
            int y=scan.nextInt();
            int px=find(x),py=find(y);
            if(x>n || y>n) res++;
            else{
　　　　　　　　　　//如果是同一类动物　　
                if(D==1){
                    if(px==py && (d[x]-d[y])%3!=0) res++;//如果两个动物在一个集合中，那么它们到根节点的距离模3肯定相等，即差值模3等于0，不等于0就是假话
                    else if(px!=py){//如果不在一个集合中 见图1
                        p[px]=py;
                        d[px]=d[y]-d[x];
                    }
                }
　　　　　　　　　 //如果不是同一类动物
                else{
                    if(px==py && (d[x]-d[y]-1)%3!=0) res++;//如果两个动物在一个集合中，题意x吃y，那么x到根节点的距离取模肯定是比y到根节点的距离取模大1，所以，d[x]%3同余（d[y]+1）%3，即差值模3等于0
                    else if(px!=py){//如果不在一个集合中 见图2
                        p[px]=py;
                        d[px]=d[y]+1-d[x];
                    }
                }
            }
        }
        System.out.println(res);
    }
}