Description [title]
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 X Y", represents the X and Y are similar.
The second argument is "2 X Y", 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 and previous conflicts, then some really, that is a lie;
2) if the current X or Y greater than N, is lie;
3) currently eat as saying X X, is a lie.
Your task is the total number (1≤ N ≤50,000) and K words (0≤K≤100,000), output lies in the given N.
[Enter]
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]
Only one integer representing the number of lies.
[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
Read a lot of problem solutions and concluded that this approach is best understood by the following:
Unlike most methods, we disjoint-set open space three times as large. But the key is used twice after what representation. Here I introduce two concepts, predators and prey! Let x then y y eating predators is x, x prey is y. For the above three times the space, which we denote respectively: 1-n: represents who and who is the same. n + 1-2 * n: represents the set of natural enemies. 2 * n + 1-3 * n: represents the set of prey.
So this is our go back and look at both instructions:
1. Enter the x and y, x and y represent the same kind: for x and y, if they already have a relationship with food being eaten, then to lie. So we check the x and y + n, y + 2 * n is in the same set (y whether prey or predators of x) if so, ans ++, if not the merger. NATURAL must merge the three sets were combined, i.e. combined: - x and y - x + n and y + n - x + 2 * n and y + 2 * n
2. Enter the x and y, x represents eat y: At this point if it is a lie, then there are two cases: - 1.x and y is the same class - 2.y eat x 1, we only need to check the situation with respect to x y to whether in a collection. In the case of 2, we need to check and x + n y are in the same set (x y is not prey). If the two conditions are not met, then it would even edge up. Need to connect three sides: - 1.y + n and x (y is the prey x) - 2.x + 2 * n and y (y predators is x) - 3. This is the most ingenious point (!): - x + n with y + 2 * n (in the x and the y prey to predators) - ease of understanding herein, may be introduced into an animal t - t prey animal is x, the annular structure, it must be a natural enemies y right - and so x + n y + 2 * n the essentially belongs to a class, to merge! - This is the soul of this question, we must understand, do not know how to take paper painting it!
To understand the relationship between these (I admit that was a bit around), this problem is not difficult to implement. Specific implementation code:
1 #include<bits/stdc++.h> 2 using namespace std; 3 int fa[20000001],ans; 4 int read() 5 { 6 int f=1,x=0;char s=getchar(); 7 while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();} 8 while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();} 9 x*=f; 10 return x; 11 } 12 void write(int x) 13 { 14 if(x<0) 15 { 16 putchar('-'); 17 x=-x; 18 } 19 if(x>9)write(x/10); 20 putchar(x%10+'0'); 21 } 22 int find(int x) 23 { 24 if(fa[x]!=x)fa[x]=find(fa[x]); 25 return fa[x]; 26 } 27 int main() 28 { 29 int n,m,w,x,y,k,d; 30 n=read();k=read(); 31 for(int i=1;i<=3*n;i++) 32 fa[i]=i; 33 for(int i=1;i<=k;i++) 34 { 35 d=read();x=read();y=read(); 36 if(x==y&&d==2) 37 { 38 ans++;continue; 39 } 40 if(x>n||y>n) 41 { 42 ans++;continue; 43 } 44 if(d==1) 45 { 46 if(find(x)==find(y+n)||find(x)==find(y+2*n))ans++; 47 else 48 { 49 fa[find(y)]=find(x); 50 fa[find(y+n)]=find(x+n); 51 fa[find(y+2*n)]=find(x+2*n); 52 } 53 } 54 if(d==2) 55 { 56 if(find(x)==find(y)||find(x+n)==find(y)||find(y+2*n)==find(x))ans++; 57 else 58 { 59 fa[find(y+n)]=find(x); 60 fa[find(y)]=find(x+2*n); 61 fa[find(y+2*n)]=find(x+n); 62 } 63 } 64 } 65 write(ans); 66 return 0; 67 }
// Reference: Luo Gu · P2024 food chain (multi-criteria disjoint-set) - Guess_Ha's blog - CSDN blog
https://blog.csdn.net/qq_39553725/article/details/76474124