这道题要用带权并查集，感觉没听懂只能先把自己能懂得写了；
数组b[i]表示i与根节点的关系，//0 ： 同类 1：吃 2：被吃#include
r1 == r2 && b[x] != b[y]说明x和y在同一集合，但他们与根节点的关系不是一样的，说明他们不是同类，是假话
#include<set>
#include<map>
#include<algorithm>
#include<vector>
#define int long long
const int N = 1e6;
using namespace std;
char* p1, * p2, buf[100000];
#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int read()
{int x = 0, f = 1;char ch = nc();while (ch < 48 || ch>57){if (ch == '-')f = -1;ch = nc();}while (ch >= 48 && ch <= 57)x = x * 10 + ch - 48, ch = nc();return x * f;
}
int a[N], b[N],ans=0;
int find(int x) {if (a[x] == x)return x;int t = a[x];a[x] = find(a[x]);b[x] = (b[x] + b[t]) % 3;//压缩路径return a[x];
}
signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int n, k;cin >> n >> k;for (int i = 1; i <= n; i++)a[i] = i;for (int i = 1; i <= k; i++) {int op, x, y;cin >> op >> x >> y;if (op == 2 && x == y)ans++;else if (x > n || y > n)ans++;else {int r1 = find(x);int r2 = find(y);if (op == 1) {if (r1 != r2) {a[r2] = r1;b[r2] = (3 - b[y] + b[x]) % 3;}else if (r1 == r2 && b[x] != b[y])ans++;}if (op == 2) {if (r1 != r2) {a[r2] = r1;b[r2] = (3 - b[y] + b[x] + 1) % 3;}else if (r1 == r2 && (b[y] - b[x] + 3) % 3 != 1)ans++;}}}cout << ans;return 0;
}
