POJ 1182 食物链 并查集

题目链接：点击打开链接

思路：并查集判断是否有关系

类型一个三种

#include<iostream>
#include<string>
using namespace std;
const int N = 500005;
int father[N];
int sum = 0;
int d[N] = { 0 };//代表与根节点的距离，有0,1,2三种
int n;
int find(int x) {//find时更新距离
	if (x != father[x]) {
		int pre = father[x];
		father[x] = find(father[x]);
		d[x] += d[pre];
		d[x] %= 3;
	}
	return father[x];
}
int judge(int k, int x, int y) {
	int a = find(x);
	int b = find(y);
	if (a == b) {
		if (k == 1 && d[x] != d[y])return 0;
		if (k == 2 && d[x] == 0 && d[y] != 2)return 0;
		if (k == 2 && d[x] == 1 && d[y] != 0)return 0;
		if (k == 2 && d[x] == 2 && d[y] != 1)return 0;
		return 1;
	}
	else {
		father[a] = b;
		if (k == 1)d[a] = (d[y] - d[x] + 3) % 3;
		if (k == 2)d[a] = (d[y] - d[x] + 4) % 3;
		return 1;
	}
}
int main() {
	int T;
	cin >> n >> T;
	for (int i = 0; i <= n; i++)father[i] = i;
	while (T--) {
		int k, a, b;
		//cin >> k >> a >> b;
		scanf("%d %d %d", &k, &a, &b);
		if (a > n || b > n) {
			sum++;
			continue;
		}
		if (k == 2 && a == b) {
			sum++;
			continue;
		}
		int y = judge(k, a, b);
		if (y == 0)sum++;
	}
	cout << sum << endl;
}