luogu2024 food chain

Topic description

There are three groups of animals A, B, and C in the animal kingdom, and the food chains of these three groups form an interesting loop. A eats B, B

Eat C, C eat A.

There are N animals, numbered 1-N. Every animal is one of A,B,C, but we don't know

Which one is it.

Some people describe the food chain relationship formed by these N animals in two ways:

The first statement is "1 XY", which means that X and Y are of the same kind.

The second is "2 XY", which means X eats Y.

This person uses the above two statements to say K sentences sentence after sentence to N animals. Some of these K sentences are true.

Yes, some are fake. When a sentence satisfies one of the following three conditions, the sentence is false, otherwise it is true.

• The current words conflict with some previous true words and are false words

• If X or Y is greater than N in the current word, it is a lie

• The current word means X eats X, which is a lie

Your task is to output the total number of false statements given N and K sentences.

Input and output format

Input format:

Enter data from eat.in

The first line has two integers, N, K, indicating that there are N animals, K sentences.

The second line starts with one sentence per line (according to the requirements of the title, see the example)

Output format:

output to eat.out

One line, an integer, representing the total number of falsehoods.

Input and output example

Input Example #1: Copy

Output Sample #1: Copy

illustrate

1 ≤ N ≤ 5 ∗ 10^4

1 ≤ K ≤ 10^5

Ideas: The idea of this question is basically the same as that of detaining criminals. We will use three arrays (here I use an array for convenience, assuming i is itself, then i+n is his prey, and i+n*2 is his enemy) When x and y are of the same kind, we only need to judge whether y is the prey of x, and whether y is the enemy of x, if not, then this is not a lie, if they are directly merged, their enemy Merged, their prey merged. The same can be obtained......

Code:

#include<iostream> 
#include <cstdio>
 #define X 300000+7
 using  namespace std;
 int tot,fa[X]; // fa[i] refers to itself, fa[i+n] is what he can beat, fa[i+n*2] is to eat his 
int read()
{
    int sum=0;
    char ch=getchar();
    while(ch>'9'||ch<'0')ch=getchar();
    while(ch>='0'&&ch<='9')sum=sum*10+ch-48,ch=getchar();
    return sum;
}
int find(int x)
{
    return fa[x]==x?x:fa[x]=find(fa[x]);
}
int hb(int x,int y)
{
    int ff=find(x),fff=find(y);
    fa [ff] = fff;
}
intmain ()
{
    int n,k;
    n=read();
    k=read();
    for(int i=1;i<=n*3;++i)
        fa[i]=i;
    for(int i=1,x,u,v;i<=k;++i)
    {
        x=read();u=read();v=read();
        if(u>n||v>n){
            ++tot;
            continue; 
        };    
        int ff=find(v);
        if(x==1)
        {
            
            if(find(u+n)==ff||find(u+n*2)==ff){
                ++tot;
                continue;
            }
            hb(u,v);hb(u+n,v+n);hb(u+n*2,v+n*2);
        }
        else
        {
            if(u==v){
                ++tot;
                continue;
            }
            if(find(u)==ff||find(u+2*n)==ff){
                ++tot;
                continue;
            }
            hb(u,v+2*n);hb(u+n,v);hb(u+2*n,v+n);
        }
    }
    cout << tot;
}