[NOI2001] 食物链

题目链接：##

传送门

题目描述：##

扩展域并查集
三种动物会构成一个食物环，可参考石头剪刀布
由于对于一个\(x\)有三种关系：

• 同类
• 食物
• 天敌

所以将一个\(x\)拆分成三个域，然后对于两种操作分别合并同类项
对于\(x\)和\(y\)是同类的情况，直接一一对应合并两者的三个域即可。
对于\(x\)吃\(y\)的情况，将\(x\)的同类并\(y\)的天敌，\(x\)的食物并\(y\)的同类，\(x\)的天敌并\(y\)的食物

对于\(x\)和\(y\)是同类的操作，有两类关系是不合法的：

• \(x\)吃\(y\)
• \(y\)吃\(x\)
  对于\(x\)吃\(y\)的操作，有两类关系是不合法的：
• \(x\)和\(y\)是同类
• \(y\)吃\(x\)

代码：##

#include<bits/stdc++.h>
#define N 3 * (50000)
using namespace std;
inline int read() {
	int cnt = 0, f = 1; char c;
	c = getchar();
	while (!isdigit(c)) {
		if (c == '-') f = -1;
		c = getchar();
	}
	while (isdigit(c)) {
		cnt = c - '0' + cnt * 10;
		c = getchar();
	}
	return cnt * f;
}
int fa[N], n, k, cnt = 0, opr;
int x, y;
int get_father(int x) {
	if(fa[x] == x) return x;
	return fa[x] = get_father(fa[x]);
}
bool check(int opr, int x, int y) {
	if(x > n || y > n) return false;
	int x1 = get_father(x), x2 = get_father(x + n), x3 = get_father(x + 2 * n);
	int y1 = get_father(y), y2 = get_father(y + n), y3 = get_father(y + 2 * n);
	if (opr == 1) {
		if(x2 == y1) return false;
		if(x1 == y2) return false;
	}
	if (opr == 2) {
		if(x == y) return false;
		if(x1 == y1) return false;
		if(x1 == y2) return false;
	}
	return true;
}
bool merge(int opr, int x, int y) {
	int x1 = get_father(x), x2 = get_father(x + n), x3 = get_father(x + 2 * n);
	int y1 = get_father(y), y2 = get_father(y + n), y3 = get_father(y + 2 * n);
	if (!check(opr, x, y)) return false;
	if (opr == 1) {
		fa[y1] = x1; fa[y2] = x2; fa[y3] = x3;
	}
	if (opr == 2) {
		fa[y1] = x2; fa[y2] = x3; fa[y3] = x1;
	}
	return true;
}
int main() {
	n = read(); k = read();
	for (register int i = 1; i <= 3 * n; i++) fa[i] = i;
	for (register int i = 1; i <= k; i++) {
		opr = read(); x = read(); y = read();
		cnt += !merge(opr, x, y);
	}
	printf("%d", cnt);
	return 0;
}